In this set of practice questions, we will summarize the main concepts of thermochemistry such as the relationship between internal energy, work and heat, exothermic and endothermic process, heat capacity, constant pressure calorimetry, constant-volume calorimetry, the enthalpy, the standard enthalpies of formation and their use in determining the heat of the reaction, and the Hess’s law.

The links to the corresponding topics are given below:

- Energy Related to Heat and Work
- Endothermic and Exothermic Processes
- Heat Capacity and Specific Heat
- Heat Capacity Practice Problems
- What is Enthalpy
- Constant-Pressure Calorimetry
- Bomb calorimeter – Constant Volume Calorimetry
- Stoichiometry and Enthalpy of Chemical Reactions
- Hess’s Law and Enthalpy of Reaction
- Hess’s Law Practice Problems
- Standard Enthalpies of Formation
- Enthalpy of Reaction from Enthalpies of Formation

#### Practice

What is the sign of Δ*E*_{system} if energy flows from the surroundings into a chemical system?

+Δ*E*

If energy flows to the system, the internal energy is going to increase, and therefore, the sign for Δ*E* is positive.

A gas absorbs 62 kJ of heat and does 48 kJ of work. Calculate Δ*E*.

Δ*E *= 16 kJ

Δ*E *= q + w

If the heat is absorbed by the system, then it has a positive sign. The job done by the system indicates a negative sign for w.

Δ*E *= +62 kJ – 48 kJ = 16 kJ

Calculate the change in internal energy of the system if it releases 415 kJ of heat and does 562 kJ of work on the surroundings.

Δ*E *= -977 kJ

Δ*E *= q + w

If the heat is released by the system, then it has a negative sign. The job done by the system also indicates a negative sign for w.

Δ*E *= -415 kJ – 562 kJ = -977 kJ

Calculate Δ*E *for each of the following.

**a) ***q *= -62 kJ, w *= +*59 kJ

**b) ***q *= +48 kJ, w *= –*56 kJ

**c) ***q *= -92 kJ, w *= *0 kJ

**a) **Δ*E *= -3 kJ

**b) **Δ*E *= -8 kJ

** ****c) **Δ*E *= -62 kJ

**a) **Δ*E *= q + w

Δ*E *= -62 kJ + 59 kJ = -3 kJ

**b) **Δ*E *= q + w

Δ*E *= +48 *–*56 kJ = -8 kJ

** **

**c) **Δ*E *= q + w

** **Δ*E *= -92 kJ + 0 kJ = -62 kJ

The gas in a piston is heated up by absorbing 984 J of heat and expands performing 541 J of work on the surroundings.

What is the change in internal energy for the system (the gas)?

Δ*E *= 443 kJ

Δ*E *= q + w

If the heat is absorbed by the system, then it has a positive sign. The job done by the system indicates a negative sign for w.

Δ*E *= +984 kJ – 541 kJ = 443 kJ

A pump is held under 1.30 atm external pressure. How much work (in J) is required to expand its volume from 1.80 L to 3.50 L?

w = 224 J

W = -PΔV

W = -1.30 atm (3.50 L – 1.80 L) = -2.21 atm · L

The last step is to convert atm · L to Joules. 1 L· atm = 101.3 J, therefore:

\[{\rm{w}}\;{\rm{ = }}\;{\rm{ – 2}}{\rm{.21 }}\cancel{{{\rm{atm}}\;{\rm{\cdot}}\;{\rm{L}}}}\;{\rm{ \times }}\;\frac{{{\rm{101}}{\rm{.3 J}}}}{{{\rm{1}}\;\cancel{{{\rm{atm}}\;{\rm{\cdot}}\;{\rm{L}}}}}}\;{\rm{ = }}\;{\rm{224}}\;{\rm{J}}\;\]

A balloon filled with 8.50 moles of nitrogen at 23.0 ^{o}C expands from 3.54 L to 5.64 L by increasing its temperature to 58.0 ^{o}C under constant pressure of 1.10 atm. Calculate *q*, *w*, and Δ*E *in kJ for the nitrogen gas in the balloon if its heat capacity is 29.15 J/°C · mol.

a) q = 8.67 kJ

b) W = -0.234 kJ

c) Δ*E = * 8.44 kJ

a) q = n · c · ΔT

q = 8.50 mol x 29.15 J/°C · mol x (58.0 ^{o}C – 23.0 ^{o}C) = 8.67 kJ

\[{\rm{q}}\;{\rm{ = }}\;{\rm{8}}{\rm{.50}}\;\cancel{{{\rm{mol}}}}\;{\rm{ \times }}\;\frac{{{\rm{29}}{\rm{.15}}\;{\rm{J}}}}{{\cancel{{^{\rm{o}}{\rm{C}}}}\;\cancel{{{\rm{mol}}}}}}\;{\rm{x}}\;\left( {{\rm{58}}{\rm{.0}}\;\cancel{{^{\rm{o}}{\rm{C}}}}\;{\rm{ – }}\;{\rm{23}}{\rm{.0}}\;\cancel{{^{\rm{o}}{\rm{C}}}}} \right)\;{\rm{ = }}\;{\rm{8}}{\rm{.67}}\;{\rm{kJ}}\]

b) W = -PΔV

W = -1.10 atm (5.64 L – 3.54 L) = -2.31 atm · L

The last step is to convert atm · L to kJ. 1 L· atm = 101.3 J, therefore:

\[{\rm{w}}\;{\rm{ = }}\;{\rm{ – 2}}{\rm{.31 }}\cancel{{{\rm{atm}}\;{\rm{\cdot}}\;{\rm{L}}}}\;{\rm{ \times }}\;\frac{{{\rm{101}}{\rm{.3 }}\cancel{{\rm{J}}}}}{{{\rm{1}}\;\cancel{{{\rm{atm}}\;{\rm{\cdot}}\;{\rm{L}}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{kJ}}}}{{{\rm{1000}}\;\cancel{{\rm{J}}}}}\;{\rm{ = }}\; – {\rm{0}}{\rm{.234}}\;{\rm{kJ}}\;\]

c) Δ*E = *q + w

ΔE = 8.67 kJ + (-0.234 kJ) = 8.44 kJ

How much heat in kJ is required to warm 1.50 L of water from 25.0 ^{o}C to 100.0 °C? (Assume a density of 1.0 g/mL for the water.)

4.70 x 10^{2} kJ

The water doesn’t go through any phase changes so, we can use the following formula for the relationship between the amount of heat needed to raise the temperature of a given substance and the corresponding temperature increase:

** **q = m × C_{s} × ΔT

Where, q is the amount of heat in J, m is the mass, C_{s} is the specific heat capacity of the material in J/g ^{o}C , and ΔT is the change in temperature (T_{final} – T_{initial}). Remember, the specific heat is the quantity of heat energy necessary to increase the temperature of 1 g of a substance by 1°C and it is 4.18 J/g ^{o}C for water. Therefore, to find the amount of heat for a given mass, we multiply it by this mass.

Before using the numbers, make sure the units match. The C_{s }is given for a gram of water so, we need to convert the volume to mL first, and then to mass in g.

\[{\rm{V}}\;{\rm{ = }}\;{\rm{1}}{\rm{.50}}\;{\rm{L}}\;{\rm{ \times }}\;\frac{{{\rm{1,000}}\;{\rm{mL}}}}{{{\rm{1}}\;{\rm{L}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.50}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{mL}}\]

m = d x V

\[{\rm{m}}\;{\rm{ = }}\;{\rm{1}}{\rm{.50}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{mL}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{g}}}}{{{\rm{mL}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.50}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{g}}\]

And now, we can plug the numbers into the equation of the heat:

q = m × C_{s} × ΔT

\[{\rm{q}}\;{\rm{ = }}\;{\rm{1}}{\rm{.50}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{3}}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{{\rm{4}}{\rm{.18}}\;{\rm{J}}}}{{\cancel{{\rm{g}}}\;\cancel{{^{\rm{o}}{\rm{C}}}}}}\;{\rm{ \times }}\;{\rm{(100}}{\rm{.0}}\;{\rm{ – }}\;{\rm{25}}{\rm{.0)}}\;\cancel{{^{\rm{o}}{\rm{C}}}}\;{\rm{ = }}\;{\rm{4}}{\rm{.70}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{{\rm{5}}\;}}{\rm{J}}\]

To convert the J to kJ, we need to divide it by 1,000, so it is 4.70 x 10^{2} kJ

What is the final temperature when a 40 g sample of water at 90 °C is mixed with a 60 g sample at 25 °C?

T_{f }= 51 °C

The basis of solving this problem is the assumption that **q _{lost} = q_{gained}**

_{.}The heat flows from the warmer sample (q

_{lost}) to the cooler one (q

_{gained}), and the amount of heat lost by the warmer sample is equal to the amount of heat gained by the cooler sample. This means that there is no heat lost to the surroundings.

Recall that q = m · C · ΔT, where m is the mass, C is the specific heat, and ΔT is the temperature change.

We can also write this as q_{lost} + q_{gained }= 0 or

q_{1 }+ q_{2} = 0

For our problem, q_{1 }= mCΔT_{1 }, q_{2 }= mCΔT_{2 }

so,** **mCΔT_{1 }+** **mCΔT_{2} = 0

ΔT_{1 }= T_{f }– 90, and ΔT_{2 }= T_{f }– 25

Where T_{f }is the final temperature of the system, so it is the same for the worm and cold samples.

Plug in the numbers and solve for T_{f }in the following equation:

40 g x C x (T_{f }– 90) + 60 g x C x (T_{f }– 25) = 0

Because C (heat capacity of water) appears on both sides of the equation and can be canceled out.

40 g x (T_{f }– 90) + 60 g x (T_{f }– 25) = 0

T_{f }= 51 °C

Notice that the equation can also be set up by using kelvin (K).

How much heat does it take to increase the temperature of a 540.6-g sample of Fe from 20.0 °C to 84.3 °C? The specific heat of iron = 0.450 J/g °C.

15.6 kJ

A similar question to that in problem 8, and we are going to use the following equation:

** **q = m × C_{s} × ΔT

Where, q is the amount of heat in J, m is the mass, C_{s} is the specific heat capacity of the material in J/g ^{o}C , and ΔT is the change in temperature (T_{final} – T_{initial}). Remember, the specific heat is the quantity of heat energy necessary to increase the temperature of 1 g of a substance by 1°C.

All we need to do is to plug the numbers into the equation of the heat:

\[{\rm{q}}\;{\rm{ = }}\;{\rm{540}}{\rm{.6}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{{\rm{0}}{\rm{.450}}\;{\rm{J}}}}{{\cancel{{\rm{g}}}\;\cancel{{^{\rm{o}}{\rm{C}}}}}}\;{\rm{ \times }}\;{\rm{(84}}{\rm{.3}}\;{\rm{ – }}\;{\rm{20}}{\rm{.0)}}\;\cancel{{^{\rm{o}}{\rm{C}}}}\;{\rm{ = }}\;{\rm{15}}{\rm{.6}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{{\rm{3}}\;}}{\rm{J}}\]

To convert the J to kJ, we need to divide it by 1,000, so it is 15.6 kJ

Calculate the specific heat capacity of a metal if a 17.0 g sample requires 481 J to change the temperature of the metal from 25.0 °C to 67.0 °C?

0.674 J/g°C

We are going to rearrange the following equation for the relation of heat, specific heat capacity, and temperature change:

** **q = m × C × ΔT

\[{\rm{C}}\;{\rm{ = }}\;\frac{{\rm{q}}}{{{\rm{m}}\;{\rm{\Delta T}}}}\]

\[{\rm{C}}\;{\rm{ = }}\;\frac{{{\rm{481}}\;{\rm{J}}}}{{{\rm{17}}{\rm{.0}}\;{\rm{g}}\;{\rm{(67}}{\rm{.0}}\;{\rm{ – }}\;{\rm{25}}{\rm{.0)}}{\;^{\rm{o}}}{\rm{C}}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.674}}\;{\rm{J}}}}{{{\rm{g}}{\;^{\rm{o}}}{\rm{C}}}}\;\]

Calculate the energy of combustion for one mole of butane if burning a 0.367 g sample of butane (C_{4}H_{10}) has increased the temperature of a bomb calorimeter by 7.73 °C. The heat capacity of the bomb calorimeter is 2.36 kJ/ °C.

2.88 x 10^{3} kJ

Because the combustion of butane increased the temperature of a bomb calorimeter, it is an exothermic reaction. This means, the heat goes from the reaction to the calorimeter and assuming there is no heat loss, the heat released by the reaction is equal to the heat absorbed by the calorimeter:

-q_{reaction} = q_{cal}

Since the reaction occurs under conditions of constant volume, q_{rxn} = ΔE_{rxn, }so we need to calculate the q_{cal }using the following formula:

q_{cal} = C_{cal} x ΔT

q_{cal} = 2.36 kJ/ °C x 7.73 °C = 18.2 kJ

Therefore, q_{rxn }= -18.2 kJ

This is the change in the internal energy of the reaction for that specific amount of butane that was burned.

To get ΔE_{rxn} per mole of butane, we need to divide q_{rxn} by the number of moles that actually reacted. To find the number of moles, we use the mas and molar mass of butane:

n(C_{4}H_{10}) = 0.367 g/58.0 g/mol = 0.00633 mol

And after this, we divide the heat of the reaction by the number of moles of butane, to get the heat of the reaction per mole of butane:

ΔE_{rxn }= 18.2 kJ/0.00633 mol = 2.88 x 10^{3} kJ/mol

Alternatively, we can set up a cross-multiplication correlation:

0.00633 mol C_{4}H_{10 }– 18.2 kJ

1 mol C_{4}H_{10 }– X kJ

X = 2.88 x 10^{3} kJ/mol

How many joules of energy is required to melt 40.0 g of ice at 0 °C? The heat of fusion (Δ*H*_{fus}) for ice is 334.0 J/g.

1.34 x 10^{3} J

Since the ice/water is already at 0 ^{o}C (melting/freezing temperature), the added heat does not change the temperature of the ice and water mixture, but rather it is used for the transition from solid to liquid.

To calculate the amount of energy required to achieve a state change, in this case melting, of the substance that is already at the state-change temperature, we use the following formula:

q = mΔ*H*_{fus}

where q is the quantity of heat energy, m is the mass, and Δ*H*_{fus} is the enthalpy of fusion, sometimes called the heat of fusion. It is the energy required to fuse (melt) a gram (or a mole) of a substance.

Plugging the numbers in the formula, we get:

q = 40.0 g x 334.0 J/g = 13,360 J

Rounding off to three significant figures, we have q = 1.34 x 10^{3} J

How many kJ of energy does it take to change 36.0 g of ice at -15.0 °C to water at 0. °C ? The specific heat of ice is 2.10 J/g°C and the heat of fusion (Δ*H*_{fus}) for ice is 334.0 J/g. Ignore the significant figures for this problem.

1.32 x 10^{4} J

We mentioned in the previous problem, that since the ice/water is already at 0 ^{o}C (melting/freezing temperature), the added heat does not change the temperature of the ice and water mixture, but rather it is used for the transition from solid to liquid. This is not the case here because the ice is at -15.0 oC and before melting, it needs to be first warmed to 0 ^{o}C. Therefore, there are two stages in this process; 1) heating the ice to 0 ^{o}C, 2) melting the ice.

The overall heat of this process then is:

q = q_{heating} + q_{melting}

q_{heating }is calculated by the formula we have already used several times:

q_{heating }= m × C × ΔT

\[{{\rm{q}}_{{\rm{heating}}}}\;{\rm{ = }}\;{\rm{36}}{\rm{.0}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{2.10\;{\rm{J}}}}{{\cancel{{\rm{g}}}\;\cancel{{^{\rm{o}}{\rm{C}}}}}}\;{\rm{ \times }}\;{\rm{(0}}\;{\rm{ – }}\;( – {\rm{15}}{\rm{.0)}}\;\cancel{{^{\rm{o}}{\rm{C}}}}\;{\rm{ = }}\;{\rm{1,134}}\;{\rm{J}}\]

To calculate the amount of energy required to melt the ice, we use the following formula:

q = mΔ*H*_{fus}

\[{{\rm{q}}_{{\rm{melting}}}}\;{\rm{ = }}\;{\rm{36}}{\rm{.0}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{334.0\;{\rm{J}}}}{{\cancel{{\rm{g}}}\;}}\;{\rm{ = }}\;{\rm{12,024}}\;{\rm{J}}\]

q = 1134 J + 12,024 J = 13,158 J

Rounding of to three significant figures, we have q = 1.32 x 10^{4} J or 13.2 kJ

The enthalpy change for the reaction is given below:

2CH_{3}OH(*l*) + 3O_{2}(*g*) → 4H_{2}O(*l*) + 2CO_{2}(*g*) Δ*H =* -1452.8 kJ

**a)** What quantity of heat is released for each mole of water formed?

**b) **What quantity of heat is released for each mole of oxygen reacted?

**a)** 363.2 kJ

**b)** 484.27 kJ

**a)** The Δ*H* value for a reaction is given specifically **based on the coefficients in the balanced equation**. Because the coefficient in front of H_{2}O is a 4, the formation of one mole of water will release:

-1452.8 kJ ÷ 4 = 363.2 kJ

**b)** According to the equation, when 3 moles of oxygen react, 8 kJ energy is released. Therefore, for the reaction of one-mole oxygen,

Δ*H = *-1452.8 kJ ÷ 3 = 484.27 kJ

How much heat will be released if 44.8 g of SO_{2} is reacted with an excess of oxygen according to the following chemical equation?

2SO2(*g*) + O_{2}(*g*) → 2SO_{3}(*g*), Δ*H*° = –198 kJ

69.2 kJ

The Δ*H* value for a reaction is given specifically **based on the coefficients in the balanced equation**. So, in this reaction, the combustion of two moles of SO_{2 }produces 198 kJ of heat. What we need to do here is first convert the mass of SO_{2 }to moles, and then calculate the amount of heat that will be released from this amount considering that burning two moles of SO_{2 }gives 198 kJ of heat.

n(SO_{2}) = 44.8 g/64.1 g/mol = 0.699 mol

Let’s now set up a cross multiplication, which would read as “2 mol SO_{2 }gives 198 kJ heat, 0.699 mol SO_{2 }will give an unknown amount of heat:

2 mol SO_{2 }– 198 kJ

0.699 mol SO_{2 }– X kJ

X = 69.2 kJ

Alternatively, we can do this by unit conversion method:

\[\Delta H\; = \;\frac{{{\rm{198}}\;{\rm{kJ}}}}{{\cancel{{{\rm{2}}\;{\rm{mol}}\;{\rm{S}}{{\rm{O}}_{\rm{2}}}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.699 }}\cancel{{{\rm{mol}}\;{\rm{S}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ = }}\;{\rm{69}}{\rm{.2}}\;{\rm{kJ}}\]

So, 69.2 kJ heat will be released if 44.8 g (0.699 mol ) of SO_{2 }is reacted according to the given chemical equation.

What is Δ*H*° for the following reaction

2C_{6}H_{6}(*l*) + 15O_{2}(*g*) → 12CO_{2}(*g*) + 6H_{2}O(*l*), Δ*H*° = ? kJ

if the consumption of 27.3 g of benzene (C_{6}H_{6}) produces 1144 kJ of heat?

6.54 x 10^{3} kJ

The Δ*H* value for a reaction is given specifically **based on the coefficients in the balanced equation**. So, in this reaction, we need to determine the combustion of two moles of C_{6}H_{6.}

Convert the mass of C_{6}H_{6}_{ }to moles, and then calculate the amount of heat that will be released from 2 moles of C_{6}H_{6} considering that burning 27.3 g C_{6}H_{6 }gives 1144 kJ of heat.

n(C_{6}H_{6}) = 27.3 g/78.1 g/mol = 0.350 mol

Let’s now set up a cross multiplication, which would read as “0.350 mol C_{6}H_{6}_{ }gives 1144 kJ heat, 2 mol C_{6}H_{6}_{ }will give an unknown amount of heat:

0.350 mol C_{6}H_{6}_{ }– 1144 kJ

2 mol C_{6}H_{6}_{ }– X kJ

X = 6.54 x 10^{3} kJ

Alternatively, we can do this by unit conversion method:

\[\Delta H\; = \;\frac{{{\rm{1144}}\;{\rm{kJ}}}}{{\cancel{{{\rm{0}}{\rm{.350}}\;{\rm{mol}}\;{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{6}}}}}}}\;{\rm{ \times }}\;{\rm{2}}\cancel{{{\rm{mol}}\;{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{\rm{6}}}}}\;{\rm{ = }}\;{\rm{6}}{\rm{.54}}\;{\rm{ \times }}\;{10^3}\;{\rm{kJ}}\]

Based on the heat of reaction for the chlorination of methane, how much heat will be released if 233.6 grams of hydrochloric acid are formed?

CH_{4}(*g*) + 3Cl_{2}(*g*) → CHCl_{3}(*l*) + 3HCl(*g*), Δ*H*° = -334 kJ

713 kJ

The Δ*H* value for a reaction is given specifically **based on the coefficients in the balanced equation **which imply their mole numbers. Therefore, converting the mass to moles is always going to be a good idea.

n(HCl) = 233.6/36.5 g/mol = 6.40 mol

According to the chemical equation, for every three moles of HCl forming, there is 334 kJ heat released. So, we need to find how much heat will be released if 6.40 mol HCl is formed.

We can set up a cross multiplication, which would read as “3 mol HCl_{ }gives 334 kJ heat, 6.40 mol HCl_{ }will give an unknown amount of heat:

3 mol HCl_{ }– 334 kJ

6.40 mol HCl_{ }– X kJ

X = 713 kJ

Alternatively, we can do this by unit conversion method:

\[\Delta H\; = \;\frac{{{\rm{334}}\;{\rm{kJ}}}}{{{\rm{3 }}\cancel{{{\rm{mol HCl}}}}}}\;{\rm{ \times }}\;{\rm{6}}{\rm{.40}}\cancel{{{\rm{mol}}\;{\rm{HCl}}}}\;{\rm{ = }}\;713\;{\rm{kJ}}\]

Calculate how many kJ of heat-energy will be released when 12.65 g of magnesium carbonate reacts with 650. mL of 0.400* M* hydrochloric acid?

MgCO_{3}(*s*)* *+* *2HCl(*aq*)* *→* *MgCl_{2}(*aq*)* *+* *H_{2}O(*l*)* *+* *CO_{2}(*g*), Δ*H*°* *=* *–112* *kJ

14.6 kJ

First, let’s convert all the quantities to moles.

n(MgCO_{3}) = 12.65 g/84.3 g/mol = 0.150 mol

The moles of HCl are calculated using the equation for molarity:

M = n/V

n = MV

n(HCl) = MV = 0.400 M x 0.650 L = 0.260 mol

We have the moles of both reactants, and therefore, we need to find the limiting reactant in order to do the calculations for the heat of the reaction. Remember, the limiting reactant is the one that gives less product. So, to find the LR, we determine whether the MgCO_{3}_{ }or HCl could produce less product based on their moles and the reaction stoichiometry. Let’s do the calculations based on the number of MgCl_{2} moles formed:

\[{\rm{n}}\left( {{\rm{MgC}}{{\rm{l}}_{\rm{2}}}\;{\rm{from}}\;{\rm{MgC}}{{\rm{O}}_{\rm{3}}}} \right){\rm{\; = }}\;{\rm{0}}{\rm{.150}}\;\cancel{{{\rm{mol}}\;{\rm{MgC}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{MgC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{MgC}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.150}}\;{\rm{mol}}\]

\[{\rm{n}}\left( {{\rm{MgC}}{{\rm{l}}_{\rm{2}}}\;{\rm{from}}\;{\rm{HCl}}} \right){\rm{\; = }}\;{\rm{0}}{\rm{.260}}\;\cancel{{{\rm{mol}}\;{\rm{HCl}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{MgC}}{{\rm{l}}_{\rm{2}}}}}{{{\rm{2}}\;\cancel{{{\rm{mol}}\;{\rm{HCl}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.130}}\;{\rm{mol}}\]

So, HCl gives less MgCl_{2}, therefore, it is the limiting reactant, and we need to calculate the amoiunt of heat beased on 0.260 mol HCl.

We can set up a cross multiplication, which would read as “2 mol HCl_{ }gives 112 kJ heat, 0.260 mol HCl_{ }will give an unknown amount of heat:

2 mol HCl_{ }– 112 kJ

0.260 mol HCl_{ }– X kJ

X = 14.6 kJ

Alternatively, we can do this by unit conversion method:

\[\Delta H\; = \;\frac{{{\rm{112}}\;{\rm{kJ}}}}{{{\rm{2 }}\cancel{{{\rm{mol}}\;{\rm{HCl}}}}}}\;{\rm{ \times }}\;{\rm{0}}{\rm{.260}}\cancel{{{\rm{mol}}\;{\rm{HCl}}}}\;{\rm{ = }}\;14.6\;{\rm{kJ}}\]

Using the standard heats of formation given below, calculate the heat of reaction for the combustion of ammonia:

4NH_{3}(*g*) + 5O_{2}(*g*) → 4NO(*g*) + 6H_{2}O(*g*), Δ*H*°_{rxn} = ?

Δ*H*_{f}° for NH_{3}(*g*) = –46.2 kJ/mol

Δ*H*_{f}° for NO(*g*) = 90.4 kJ/mol

Δ*H*_{f}° for H_{2}O(*g*) = –241.8 kJ/mol

Δ*H*°_{rxn} = -904.4 kJ

To calculate 𝚫*H*°_{rxn}, we need to subtract the enthalpies of formation of the reactants multiplied by their stoichiometric coefficients from the enthalpies of formation of products multiplied by their stoichiometric coefficients:

Δ*H*°_{rxn} = Σ*n*_{p}Δ*H*^{o}_{f} (products) – Σ*n*_{r}Δ*H*°_{f }(reactants)

Where n_{p} and n_{r} are the molar coefficients of the products and reactants in the balanced chemical equation.

Remember, the standard enthalpy for the formation of elements or their molecules in standard states is, by definition, equal to zero.

We can now enter the numbers to determine the heat of our reaction:

Δ*H*°_{rxn} = [4 x Δ*H*^{o}_{f} (NO) + 6 x Δ*H*^{o}_{f} (H_{2}O)] – [4 x Δ*H*^{o}_{f} (NH_{3}) + 5 x Δ*H*^{o}_{f} (O_{2})]

Δ*H*°_{rxn} = [4 x 90.4 kJ + 6 x (–241.8)] – [4 x (–46.2 kJ) + 5 x 0]

Δ*H*°_{rxn} = -904.4 kJ

Δ*H*°_{rxn }is negative which means it is an exothermic reaction releasing 904.40kJ of heat.

Combustion of butane (C_{4}H_{10}) releases 5755 kJ of energy according to the following chemical equation.

2C_{4}H_{10}(*g*) + 13O_{2}(*g*) → 8CO_{2}(*g*) + 10H_{2}O(*l*), Δ*H*°_{rxn} = -5755 kJ

Calculate the molar enthalpy of formation of butane using the information given below:

Δ*H*_{f}° for CO_{2}(*g*) = –393.5 kJ/mol

Δ*H*_{f}° for H_{2}O(*l*) = –285.8 kJ/mol

-126 kJ/mol

To calculate Δ*H*^{o}_{f} (C_{4}H_{10}), we are going to use the equation for the heat of reaction based on the standard enthalpies of formation:

Δ*H*°_{rxn} = Σ*n*_{p}Δ*H*^{o}_{f} (products) – Σ*n*_{r}Δ*H*°_{f }(reactants)

Where n_{p} and n_{r} are the molar coefficients of the products and reactants in the balanced chemical equation.

Remember, the standard enthalpy for the formation of elements or their molecules in standard states is, by definition, equal to zero.

We can now enter the numbers to determine the heat of our reaction:

Let’s add the components for our reaction:

Δ*H*°_{rxn} = [8 x Δ*H*^{o}_{f} (CO_{2}) + 10 x Δ*H*^{o}_{f} (H_{2}O)] – [2 x Δ*H*^{o}_{f} (C_{4}H_{10}) + 13 x Δ*H*^{o}_{f} (O_{2})] = -5755 kJ

Δ*H*^{o}_{f} (C_{4}H_{10}) is the unknown that we need to solve for in this equation:

Δ*H*°_{rxn} = [8 x (–393.5 kJ) + 10 x Δ*H*^{o}_{f} (–285.8 kJ)] – [2 x Δ*H*^{o}_{f} (C_{4}H_{10}) + 13 x 0] = -5755 kJ

-3148 kJ – 2858 kJ – 2 Δ*H*^{o}_{f} (C_{4}H_{10}) = -5755 kJ

-3148 kJ – 2858 kJ + 5755 kJ = 2 Δ*H*^{o}_{f} (C_{4}H_{10})

Δ*H*^{o}_{f} (C_{4}H_{10}) = -125.5 kJ/mol

Rounding off to three significant figures, we get -126 kJ/mol.

Zinc is recovered from ZnS by first oxidizing it to ZnO. Calculate the enthalpy of this oxidation reaction using the data given below:

2ZnS(*s*) + 3O_{2}(*g*) → 2ZnO(*s*) + 2SO_{2}(*g*)

Δ*H*_{f}° for SO_{2}(*g*) = -296.8 kJ/mol

Δ*H*_{f}° for ZnS(*s*) = -206.0 kJ/mol

Δ*H*_{f}° for ZnO(*s*) = -350.5 kJ/mol

-883 kJ

To calculate 𝚫*H*°_{rxn}, we need to subtract the enthalpies of formation of the reactants multiplied by their stoichiometric coefficients from the enthalpies of formation of products multiplied by their stoichiometric coefficients:

Δ*H*°_{rxn} = Σ*n*_{p}Δ*H*^{o}_{f} (products) – Σ*n*_{r}Δ*H*°_{f }(reactants)

Where n_{p} and n_{r} are the molar coefficients of the products and reactants in the balanced chemical equation.

Remember, the standard enthalpy for the formation of elements or their molecules in standard states is, by definition, equal to zero.

We can now enter the numbers to determine the heat of our reaction:

Δ*H*°_{rxn} = [2 x Δ*H*^{o}_{f} (ZnO) + 2 x Δ*H*^{o}_{f} (SO_{2})] – [2 x Δ*H*^{o}_{f} (ZnS) + 3 x Δ*H*^{o}_{f} (O_{2})]

Δ*H*°_{rxn} = [2 x (-350.5 kJ) + 2 x (-296.8 kJ)] – [2 x (-206.0 kJ) + 3 x 0]

Δ*H*°_{rxn} = -701 kJ – 593.6 kJ + 412 kJ = -883 kJ

Δ*H*°_{rxn }is negative which means it is an exothermic reaction releasing 883 kJ of heat.

Using the standard heats of formation given below, calculate the heat of reaction between barium carbonate (BaCO_{3}) and sulfuric acid (H_{2}SO_{4}).

BaCO_{3}(*s*) + H_{2}SO_{4}(*aq*) → BaSO_{4}(*s*) + CO_{2}(*g*) + H_{2}O(*l*)

Δ*H*_{f}° for BaSO_{4}(*s*) = -1473.2 kJ/mol

Δ*H*_{f}° for CO_{2}(*g*) = –393.5 kJ/mol

Δ*H*_{f}° for H_{2}O(*l*) = –285.8 kJ/mol

Δ*H*_{f}° for BaCO_{3}(*s*) = -1213.0 kJ/mol

Δ*H*_{f}° for 3H_{2}SO_{4}(*aq*) = -814.0kJ/mol

-126 kJ

To calculate 𝚫*H*°_{rxn}, we need to subtract the enthalpies of formation of the reactants multiplied by their stoichiometric coefficients from the enthalpies of formation of products multiplied by their stoichiometric coefficients:

Δ*H*°_{rxn} = Σ*n*_{p}Δ*H*^{o}_{f} (products) – Σ*n*_{r}Δ*H*°_{f }(reactants)

_{p} and n_{r} are the molar coefficients of the products and reactants in the balanced chemical equation.

We can now enter the numbers to determine the heat of our reaction:

Δ*H*°_{rxn} = [1 x Δ*H*^{o}_{f} (BaSO_{4}(*s*)) + 1 x Δ*H*^{o}_{f} (CO_{2}(*g*)) + 1 x Δ*H*^{o}_{f} (H_{2}O(*l*))] – [1 x Δ*H*^{o}_{f} (BaCO_{3}(*s*)) + 1 x Δ*H*^{o}_{f} (H_{2}SO_{4}(*aq*))]

Δ*H*°_{rxn} = [-1473.2 kJ –393.5 kJ –285.8 kJ] – [-1213.0 kJ -814.0kJ]

Δ*H*°_{rxn} = -2,152.5 kJ + 2,027 kJ = -125.5 kj

Δ*H*°_{rxn }is negative which means it is an exothermic reaction releasing 126 kJ of heat.

The heat of reaction for the reaction of nitrogen oxide and chlorine is –76 kJ:

2NO(*g*) + Cl_{2}(*g*) ⇆ 2NOCl(*g*)

What is the enthalpy change for the following reaction?

2NOCl(*g*) ⇆ 2NO(*g*) + Cl_{2}(*g*)

+76 kJ

The second equation has been obtained by reversing the first equation. Therefore, to obtain Δ*H *for the second equation, the Δ*H *for the first equation must be reversed in sign, and thus, it is +76 kJ.

The heat of reaction for the oxidation of hydrogen sulfide to sulfur dioxide accord according to the following equation is -1036 kJ:

2H_{2}S(*g*) + 3O_{2}(*g*) → 2SO_{2}(*g*) + 2H_{2}O(*g*)

What is the enthalpy change for the following reaction?

H_{2}S(*g*) + 1.5O_{2}(*g*) → SO_{2}(*g*) + H_{2}O(*g*)

-518 kJ

The second equation has been obtained by dividing the first equation by two. Therefore, to obtain Δ*H *for the second equation, the Δ*H *for the first equation must be divided by two, and thus, it is -1036 kJ ÷ 2 = -518 kJ.

Consider the thermochemical equation for the combustion of nitromethane (CH_{3}NO_{2}):

2CH_{3}NO_{2}(*l*) + 3/2O_{2}(*g*) → 2CO_{2}(*g*) + 3H_{2}O(*l*) + N_{2}(*g*), Δ*H *°_{rxn} = -1418 kJ

** **What is the enthalpy change for this reaction represented by the following equation?

** **4CH_{3}NO_{2}(*l*) + 3O_{2}(*g*) → 4CO_{2}(*g*) + 6H_{2}O(*l*) + N_{2}(*g*), **Δ H °_{rxn} = ?**

-2836 kJ

The second equation has been obtained by multiplying the first equation by two. Therefore, to obtain Δ*H *for the second equation, the Δ*H *for the first equation must be multiplied by two, and thus, it is -1418 kJ x 2 = -2836 kJ.

Using the Hess’s law, calculate Δ*H*^{o} for the reaction:

2NO_{2 }+ Cl_{2}(*g*) → 2NOCl(*g*) + O_{2}(*g*), **Δ***H*^{o}**= ?**

Use the following reactions and given Δ*H*’s:

1) 2NO(*g*) + Cl_{2}(*g*) → 2NOCl(*g*), Δ*H*^{o }= -76 kJ

2) 2NO(*g*) + O_{2}(*g*) → 2NO_{2}(*g*), Δ*H*^{o }= -114 kJ

38 kJ

NO_{2 }appears in the second reaction as a product. However, it is a reactant in the target reaction, therefore, the second reaction must be reversed before adding the two equations.

We will label the reversed equation as 2a. Remember, to change the sign of Δ*H*^{o }when reversing a reaction!

2a) 2NO_{2}(*g*) → 2NO(*g*) + O_{2}(*g*), Δ*H*^{o }= +114 kJ

And now, we can add equations 1 and 2a:

2NO(*g*) + Cl_{2}(*g*) → 2NOCl(*g*), Δ*H*^{o }= -76 kJ

+

2NO_{2}(*g*) → 2NO(*g*) + O_{2}(*g*), Δ*H*^{o }= +114 kJ

2NO(*g*) + Cl_{2}(*g*) + 2NO_{2}(*g*) → 2NOCl(*g*) + 2NO(*g*) + O_{2}(*g*)

2NO_{2}(*g*) + Cl_{2}(*g*) → 2NOCl(*g*) + O_{2}(*g*)

Δ*H*^{o }= -76 kJ +114 kJ = 38 kJ

In summary, we can write:

**2NO _{2 }+ Cl_{2}(g) → 2NOCl(g) + O_{2}(g), **

**Δ**

*H*

^{o}**= 38 kJ**

Calculate the enthalpy for the oxidation of CO to CO_{2} using the enthalpy of reaction for the combustion of C to CO (Δ*H* = -221.0 kJ) and the enthalpy for the combustion of C to CO_{2} (Δ*H* = -393.5 kJ).

2CO(g) + O_{2}(*g*) → 2CO_{2}(g) **Δ H = ?**

1) C(s) + O_{2}(*g*) → CO_{2}(g) Δ*H* = -393.5 kJ

2) 2C(s) + O_{2}(*g*) → 2CO(g) Δ*H* = -221.0 kJ

-566 kJ

The target reaction contains 2 moles of CO_{2} as a product, so we need to multiply equation 1 by two and do the same for the value of Δ*H:*

1a) 2C(s) + 2O_{2}(*g*) → 2CO_{2}(g), Δ*H* = -787 kJ

The second equation needs to be reversed since in the target reaction, CO is a reactant while here it is a product:

2a) 2CO(g) → 2C(s) + O_{2}(*g*), Δ*H* = 221.0 kJ

And now we can add the two equations:

1a) 2C(s) + 2O_{2}(*g*) → 2CO_{2}(g), Δ*H* = -787 kJ

+

2a) 2CO(g) → 2C(s) + O_{2}(*g*), Δ*H* = 221.0 kJ

2C(s) + 2O_{2}(*g*) + 2CO(g) → 2CO_{2}(g) + 2C(s) + O_{2}(*g*)

2CO(g) + O_{2}(*g*) → 2CO_{2}(g)

Δ*H* = -787 kJ + 221.0 kJ = -566 kJ

Calculate the enthalpy for the combustion reaction of sulfur to sulfur trioxide using the enthalpies of the two reactions shown below:

2S(*s*) + 3O_{2}(*g*) → 2SO_{3}(*g*) ** Δ H = ?**

1) S(*s*) + O_{2}(*g*) → SO_{2}(*g*) Δ*H* = -297 kJ

2) 2SO_{3}(*g*) → 2SO_{2}(*g*) + O_{2}(*g*) Δ*H* = 198 kJ

-792 kJ

The target reaction contains 2 moles of S as a reactant, so we need to multiply equation 1 by two and do the same for the value of Δ*H:*

1a) 2S(*s*) + 2O_{2}(*g*) → 2SO_{2}(*g*) Δ*H* = -594 kJ

The second equation needs to be reversed since in the target reaction, SO_{3} is a product while here it is a reactant:

2a) 2SO_{2}(*g*) + O_{2}(*g*) → 2SO_{3}(*g*) Δ*H* = -198 kJ

And now we can add the two equations:

1a) 2S(*s*) + 2O_{2}(*g*) → 2SO_{2}(*g*) Δ*H* = -594 kJ

+

2a) 2SO_{2}(*g*) + O_{2}(*g*) → 2SO_{3}(*g*) Δ*H* = -198 kJ

2S(*s*) + 2O_{2}(*g*) + 2SO_{2}(*g*) + O_{2}(*g*) → 2SO_{2}(*g*) + 2SO_{3}(*g*)

2S(*s*) + 3O_{2}(*g*) → 2SO_{3}(*g*)

Δ*H* = -594 kJ + (-198 kJ) = -792 kJ

Using the Hess’s law, calculate Δ*H*^{o} for the combustion reaction of butene:

C_{4}H_{8}(*g*) + 6O_{2}(*g*) → 4CO_{2}(*g*) + 4H_{2}O(*l*), **Δ***H*^{o }**= ?**

Use the following reactions and given Δ*H*’s:

1) 2H_{2}(*g*) + O_{2}(*g*) → 2H_{2}O(*g*), Δ*H*^{o }= -571 kJ

2) C_{4}H_{8}(*g*) + H_{2}(*g*) → C_{4}H_{10}(*g*), Δ*H*^{o }= -126 kJ

3) 2C_{4}H_{10}(*g*) + 13O_{2}(*g*) → 8CO_{2}(*g*) + 10H_{2}O(*l*), Δ*H*^{o }= -5754 kJ

Δ*H*^{o }= -2718 kJ

Because C_{4}H_{8 }appears as a reactant in the target reaction, we are not going to reverse equation 2. However, to get rid of C_{4}H_{10, }we need to multiply equation 2 by two since there are two moles of C_{4}H_{10 }in equation 3. Let’s label the new equation as 2a:

2a) 2C_{4}H_{8}(*g*) + 2H_{2}(*g*) → 2C_{4}H_{10}(*g*), Δ*H*^{o }= -252 kJ

** **We cannot yet add equations 1, 2a, and 3 because there is going to be H2 only in the left side of the equation, but the target equation does not contain H2. This means we need to also reverse equation 1. Notice that reversing equation 2 for the same reason puts C_{4}H_{8 }as product and that is not what we have in the target reaction.

So, let’s reverse equation 1, changing the sign of the enthalpy, and label it as 1a:

1a) 2H_{2}O(*g*) → 2H_{2}(*g*) + O_{2}(*g*), Δ*H*^{o }= 571 kJ

Now, we can add equations 1a, 2a, and 3:

1a) 2H_{2}O(*g*) → 2H_{2}(*g*) + O_{2}(*g*), Δ*H*^{o }= 571 kJ

+

2a) 2C_{4}H_{8}(*g*) + 2H_{2}(*g*) → 2C_{4}H_{10}(*g*), Δ*H*^{o }= -252 kJ

+

3) 2C_{4}H_{10}(*g*) + 13O_{2}(*g*) → 8CO_{2}(*g*) + 10H_{2}O(*l*), Δ*H*^{o }= -5754 kJ

2H_{2}O(*g*) + 2C_{4}H_{8}(*g*) + 2H_{2}(*g*) + 2C_{4}H_{10}(*g*) + __13__O_{2}(*g*) → 2H_{2}(*g*) + O_{2}(*g*) + 2C_{4}H_{10}(*g*) + 8CO_{2}(*g*) + __10__H_{2}O(*l*)

2C_{4}H_{8}(*g*) + 12O_{2}(*g*) → 8CO_{2}(*g*) + 8H_{2}O(*l*)

Δ*H*^{o }= 571 kJ + (-252 kJ) + (-5754 kJ) = -5435 kJ

The last step is to simplify the equation by dividing it by two. Remember, the same goes for the enthalpy value:

2C_{4}H_{8}(*g*) + 12O_{2}(*g*) → 8CO_{2}(*g*) + 8H_{2}O(*l*), **Δ H^{o }= -2718 kJ**

Using the Hess’s law and the enthalpies of the three combustion reactions below, calculate the enthalpy of the reaction producing methanol (CH_{3}OH) from carbon monoxide and hydrogen gas.

CO(g) + 2H_{2}(*g*) → CH_{3}OH(g), **Δ H = ?**

1) 2CO(g) + O_{2}(*g*) → 2CO_{2}(g), Δ*H* = -566 kJ

2) 2H_{2}(*g*) + O_{2}(*g*) → 2H_{2}O(*g*), Δ*H*^{o }= -571 kJ

3) 2CH_{3}OH(g) + 3O_{2}(*g*) → 2CO_{2}(g) + 4H_{2}O(*g*), Δ*H*^{o }= −1430 kJ

-139 kJ

The first two equations seem to be in the correct directions since they contain CO and H_{2 }as reactants which is what we have in the target equation.

The **third equation**, however, **must be reversed** because the methanol here is a reactant while in the target equation, it is a product. So, let’s reverse it and label it as 3a:

3a) 2CO_{2}(g) + 4H_{2}O(*g*) → 2CH_{3}OH(g) + 3O_{2}(*g*), Δ*H*^{o }= +1430 kJ

Let’s add equations 1, 2, and 3a and see what we get:

1) 2CO(g) + O_{2}(*g*) → 2CO_{2}(g), Δ*H* = -566 kJ

+

2) 2H_{2}(*g*) + O_{2}(*g*) → 2H_{2}O(*g*), Δ*H*^{o }= -571 kJ

+

3a) 2CO_{2}(g) + 4H_{2}O(*g*) → 2CH_{3}OH(g) + 3O_{2}(*g*), Δ*H*^{o }= +1430 kJ

You may notice that there are four moles of water on the left side (equation 3a), but only two moles on the right side (equation 2). This means we need to **multiply equation two** **with its enthalpy value by two**. Let’s do that and label the new equation as 2a:

2a) 4H_{2}(*g*) + 2O_{2}(*g*) → 4H_{2}O(*g*), Δ*H*^{o }= -1141 kJ

And now, we can add equations 1, 2a, and 3a:

1) 2CO(g) + O_{2}(*g*) → 2CO_{2}(g), Δ*H* = -566 kJ

+

2a) 4H_{2}(*g*) + 2O_{2}(*g*) → 4H_{2}O(*g*), Δ*H*^{o }= -1141 kJ

+

3a) 2CO_{2}(g) + 4H_{2}O(*g*) → 2CH_{3}OH(g) + 3O_{2}(*g*), Δ*H*^{o }= +1430 kJ

__________________________________________________

2CO) + O_{2} + 4H_{2} + 2O_{2} + 2CO_{2} + 4H_{2}O → 2CO_{2 }+ 4H_{2}O + 2CH_{3}OH + 3O_{2}

Cancel the same molecules on both side of the equation:

2CO + O_{2} + 4H_{2} + 2O_{2} + 2CO_{2} + 4H_{2}O → 2CO_{2} + 4H_{2}O + 2CH_{3}OH + 3O_{2}

Δ*H*^{o }= -566 kJ + (-1141 kJ) + 1430 kJ/mol = -277 kJ

So, we have:

2CO(g) + 4H_{2}(*g*) → 2CH_{3}OH(g), Δ*H*^{o }= -277 kJ

The last step is to **divide the equation by two** to match the target reaction:

CO(g) + 2H_{2}(*g*) → CH_{3}OH(g), Δ*H*^{o }= -139 kJ

Using the Hess’s law and the enthalpies of the given reactions, calculate the enthalpy of the combustion reaction of CH_{3}Cl:

CH_{3}Cl(*g*) + O_{2}(*g*) → CO(g) + HCl(*g*) + H_{2}O(*l*), **Δ H^{ }= ?**

1) CO(g) + 2H_{2}(*g*) → CH_{3}OH(g), Δ*H*^{o }= -139 kJ

2) 2H_{2}(*g*) + O_{2}(*g*) → 2H_{2}O(*g*), Δ*H*^{o }= -571 kJ

3) CH_{3}OH(*g*) + HCl(*aq*) → CH_{3}Cl(*g*) + H_{2}O(*l*), Δ*H*^{o }= -28 kJ

-404 kJ

The first reaction needs to be reversed because the CO is a reactant, while in the target reaction it is a product. Let’s reverse it and label it as 1a:

1a) CH_{3}OH(g) → CO(g) + 2H_{2}(*g*), Δ*H*^{o }= +139 kJ

The second reaction is good as it is. However, the third reaction needs to be reversed as well in order to make CH_{3}Cl as a reactant as it is in the target equation:

3a) CH_{3}Cl(*g*) + H_{2}O(*g*) → CH_{3}OH(*g*) + HCl(*aq*), Δ*H*^{o }= +28 kJ

We can now add equations 1a, 2, and 3a:

1a) CH_{3}OH(g) → CO(g) + 2H_{2}(*g*), Δ*H*^{o }= +139 kJ

+

2) 2H_{2}(*g*) + O_{2}(*g*) → 2H_{2}O(*g*), Δ*H*^{o }= -571 kJ

+

3a) CH_{3}Cl(*g*) + H_{2}O(*l*) → CH_{3}OH(*g*) + HCl(*aq*), Δ*H*^{o }= +28 kJ

____________________________________________

CH_{3}OH(g) + 2H_{2}(*g*) + O_{2}(*g*) + CH_{3}Cl(*g*) + 2H_{2}O(*l*) → CO(g) + 2H_{2}(*g*) + 2H_{2}O(*g*) + CH_{3}OH(*g*) + HCl(*aq*)

Cancel the same molecules on both side of the equation:

CH_{3}OH(g) + 2H_{2}(*g*) + O_{2}(*g*) + CH_{3}Cl(*g*) + H_{2}O(*l*) → CO(g) + 2H_{2}(*g*) + 2H_{2}O(*g*) + CH_{3}OH(*g*) + HCl(*aq*)

CH_{3}Cl(*g*) + O_{2}(*g*) → CO(g) + HCl(*g*) + H_{2}O(*g*)

Δ*H*^{o }= +139 kJ + (-571 kJ) + 28 kJ = -404 kJ

Using the Hess’s law and the enthalpies of the given reactions, calculate the enthalpy of the following oxidation reaction of ammonia:

4NH_{3}(g) + 5O_{2}(g) → 4NO(g) + 6H_{2}O(g), **Δ***H***= ?**

1) N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g), Δ*H* = -92 kJ

2) 2H_{2}(*g*) + O_{2}(*g*) → 2H_{2}O(*g*), Δ*H* = -484 kJ

3) N_{2}(g) + O_{2}(g) → 2NO(g), Δ*H* = 181 kJ

**-906 kJ**

The **first reaction needs to be reversed** because the NH_{3} is a product, while in the target reaction it is a reactant. Let’s reverse it and label it as 1a:

1a) 2NH_{3}(g) → N_{2}(g) + 3H_{2}(g), Δ*H* = +92 kJ

Let’s put the three equations together and see what the sum would look like:

1a) 2NH_{3}(g) → N_{2}(g) + 3H_{2}(g), Δ*H* = +92 kJ

2) 2H_{2}(*g*) + O_{2}(*g*) → 2H_{2}O(*g*), Δ*H* = -484 kJ

3) N_{2}(g) + O_{2}(g) → 2NO(g), Δ*H* = 181 kJ

One problem is that the hydrogens in equation 1a and 2 do not cancel each other since we have 3 moles in equation 1a and 2 moles in equation 2. Therefore, **let’s multiply equation 1a by two**, and **multiply equation 2 by three**:

1b) 4NH_{3}(g) → 2N_{2}(g) + 6H_{2}(g), Δ*H* = +184 kJ

2a) 6H_{2}(*g*) + 3O_{2}(*g*) → 6H_{2}O(*g*), Δ*H* = -1452 kJ

And now let’s put these three equations together:

1b) 4NH_{3}(g) → 2N_{2}(g) + 6H_{2}(g), Δ*H* = +184 kJ

2a) 6H_{2}(*g*) + 3O_{2}(*g*) → 6H_{2}O(*g*), Δ*H* = -1452 kJ

3) N_{2}(g) + O_{2}(g) → 2NO(g), Δ*H* = 181 kJ

The problem now is the number of moles of nitrogen in equations 1b and 3. We need to **multiply equation 3 by two:**

3a) 2N_{2}(g) + 2O_{2}(g) → 4NO(g), Δ*H* = 362 kJ

And now we can add equations 1b, 2a, and 3a:

1b) 4NH_{3}(g) → 2N_{2}(g) + 6H_{2}(g), Δ*H* = +184 kJ

+

2a) 6H_{2}(*g*) + 3O_{2}(*g*) → 6H_{2}O(*g*), Δ*H* = -1452 kJ

+

3a) 2N_{2}(g) + 2O_{2}(g) → 4NO(g), Δ*H* = 362 kJ

____________________________________________

4NH_{3}(g) + 6H_{2}(*g*) + 3O_{2}(*g*) + 2N_{2}(g) + 2O_{2}(g) → 2N_{2}(g) + 6H_{2}(g) + 6H_{2}O(*g*) + 2NO(g)

4NH_{3}(g) + 5O_{2}(*g*) → 6H_{2}O(*g*) + 2NO(g)

Δ*H* = +184 kJ + (-1452 kJ) + 362 kJ = **-906 kJ**

Using the Hess’s law and the enthalpies of the given reactions, calculate the enthalpy of the following oxidation reaction between CuO and HCl:

2CuO(s) + 4HCl(g) → 2CuCl(s) + Cl_{2}(g) + 2H_{2}O(g), **Δ***H***= ?**

1) CuO(s) + H_{2}(g) → Cu(s) + H_{2}O(g), Δ*H* = -85 kJ

2) 2Cu(s) + Cl_{2}(g) → 2CuCl(s), Δ*H* = -274 kJ

3) H_{2}(g) + Cl_{2}(g) → 2HCl(g), Δ*H* = -184 kJ

**-76 kJ**

We need to **multiply equation 1 by two** since there are two moles of CuO(s) in the target equation. Let’s label the new equation as 1a:

1a) 2CuO(s) + 2H_{2}(g) → 2Cu(s) + 2H_{2}O(g), Δ*H* = -170 kJ

Equation 2 seems to be fine, so let’s leave it as it is.

There are 2H_{2}(g) in equation 1a, and this means that we need to **reverse equation 3** and **multiply it by two** so that we can have 2H_{2}(g) as a product and cancel them with the ones in equation 1a:

3a) 4HCl(g) → 2H_{2}(g) + 2Cl_{2}(g), Δ*H* = 368 kJ

** **Let’s add equations 1a, 2, and 3a and see if they add up:

1a) 2CuO(s) + 2H_{2}(g) → 2Cu(s) + 2H_{2}O(g), Δ*H* = -170 kJ

2) 2Cu(s) + Cl_{2}(g) → 2CuCl(s), Δ*H* = -274 kJ

3a) 4HCl(g) → 2H_{2}(g) + 2Cl_{2}(g), Δ*H* = 368 kJ

2CuO(s) + 2H_{2}(g) + 2Cu(s) + Cl_{2}(g) + 4HCl(g) → 2Cu(s) + 2H_{2}O(g) + 2CuCl(s) + 2H_{2}(g) + 2Cl_{2}(g)

2CuO(s) + 4HCl(g) → 2CuCl(s) + 2H_{2}O(g) + Cl_{2}(g)

Δ*H* = -170 kJ + (-274 kJ) + 368 kJ = **-76 kJ**