## General Chemistry

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:

#### Practice

1.

What is the sign of ΔEsystem if energy flows from the surroundings into a chemical system?

E

Solution

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

2.

A gas absorbs 62 kJ of heat and does 48 kJ of work. Calculate ΔE.

ΔE = 16 kJ

Solution

ΔE = 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.

ΔE = +62 kJ – 48 kJ = 16 kJ

3.

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.

ΔE = -977 kJ

Solution

ΔE = 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.

ΔE = -415 kJ – 562 kJ = -977 kJ

4.

Calculate ΔE 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) ΔE = -3 kJ

b) ΔE = -8 kJ

c) ΔE = -62 kJ

Solution

a) ΔE = q + w

ΔE = -62 kJ + 59 kJ = -3 kJ

b) ΔE = q + w

ΔE = +48 56 kJ = -8 kJ

c) ΔE = q + w

ΔE = -92 kJ + 0 kJ = -62 kJ

5.

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)?

ΔE = 443 kJ

Solution

ΔE = 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.

ΔE = +984 kJ – 541 kJ = 443 kJ

6.

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

Solution

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:

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

7.

A balloon filled with 8.50 moles of nitrogen at 23.0 oC expands from 3.54 L to 5.64 L by increasing its temperature to 58.0 oC under constant pressure of 1.10 atm. Calculate q, w, and ΔE 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) ΔE = 8.44 kJ

Solution

a) q = n · c · ΔT

q = 8.50 mol x 29.15 J/°C · mol x (58.0 oC – 23.0 oC) = 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) ΔE = q + w

ΔE = 8.67 kJ + (-0.234 kJ) = 8.44 kJ

8.

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

4.70 x 102 kJ

Solution

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 × Cs × ΔT

Where, q is the amount of heat in J, m is the mass, Cs is the specific heat capacity of the material in J/g oC , and ΔT is the change in temperature (Tfinal – Tinitial). 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 oC 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 Cs 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 × Cs × 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 102 kJ

9.

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?

Tf  = 51 °C

Solution

The basis of solving this problem is the assumption that qlost = qgained. The heat flows from the warmer sample (qlost) to the cooler one (qgained), 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 qlost + qgained = 0 or

q+ q2 = 0

For our problem, q= mCΔT, q= mCΔT2

so, mCΔT1  + mCΔT2 = 0

ΔT= T– 90, and ΔT= T– 25

Where Tf  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 Tf  in the following equation:

40 g x C x (T– 90) + 60 g x C x (T– 25) = 0

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

40 g x (T– 90) + 60 g x (T– 25) = 0

Tf  = 51 °C

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

10.

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

Solution

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

q = m × Cs × ΔT

Where, q is the amount of heat in J, m is the mass, Cs is the specific heat capacity of the material in J/g oC , and ΔT is the change in temperature (Tfinal – Tinitial). 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

11.

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

Solution

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

q = m × C × ΔT

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

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

12.

Calculate the energy of combustion for one mole of butane if burning a 0.367 g sample of butane (C4H10) 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 103 kJ

Solution

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:

-qreaction  = qcal

Since the reaction occurs under conditions of constant volume, qrxn = ΔErxn, so we need to calculate the qcal using the following formula:

qcal = Ccal x ΔT

qcal = 2.36 kJ/ °C x 7.73 °C = 18.2 kJ

Therefore, qrxn = -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 ΔErxn per mole of butane, we need to divide qrxn by the number of moles that actually reacted. To find the number of moles, we use the mas and molar mass of butane:

n(C4H10) = 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:

ΔErxn = 18.2 kJ/0.00633 mol = 2.88 x 103 kJ/mol

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

0.00633 mol C4H10 – 18.2 kJ

1 mol C4H10 – X kJ

X = 2.88 x 103 kJ/mol

13.

How many joules of energy is required to melt 40.0 g of ice at 0 °C? The heat of fusion (ΔHfus) for ice is 334.0 J/g.

1.34 x 103 J

Solution

Since the ice/water is already at 0 oC (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 = mΔHfus

where q is the quantity of heat energy, m is the mass, and ΔHfus 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 103 J

14.

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 (ΔHfus) for ice is 334.0 J/g. Ignore the significant figures for this problem.

1.32 x 104 J

Solution

We mentioned in the previous problem, that since the ice/water is already at 0 oC (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 oC. Therefore, there are two stages in this process; 1) heating the ice to 0 oC, 2) melting the ice.

The overall heat of this process then is:

q = qheating + qmelting

qheating is calculated by the formula we have already used several times:

qheating = 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 = mΔHfus

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

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

Rounding of to three significant figures, we have q = 1.32 x 104 J or 13.2 kJ

15.

The enthalpy change for the reaction is given below:

2CH3OH(l) + 3O2(g) → 4H2O(l) + 2CO2(g) ΔH = -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

Solution

a) The ΔH value for a reaction is given specifically based on the coefficients in the balanced equation. Because the coefficient in front of H2O 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,

ΔH = -1452.8 kJ ÷ 3 = 484.27 kJ

16.

How much heat will be released if 44.8 g of SO2 is reacted with an excess of oxygen according to the following chemical equation?

2SO2(g) + O2(g) → 2SO3(g), ΔH° = –198 kJ

69.2 kJ

Solution

The ΔH 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 SOproduces 198 kJ of heat. What we need to do here is first convert the mass of SO2 to moles, and then calculate the amount of heat that will be released from this amount considering that burning two moles of SO2 gives 198 kJ of heat.

n(SO2) = 44.8 g/64.1 g/mol = 0.699 mol

Let’s now set up a cross multiplication, which would read as “2 mol SOgives 198 kJ heat, 0.699 mol SOwill give an unknown amount of heat:

2 mol SO– 198 kJ

0.699 mol SO– 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 SOis reacted according to the given chemical equation.

17.

What is ΔH° for the following reaction

2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l), ΔH° = ? kJ

if the consumption of 27.3 g of benzene (C6H6) produces 1144 kJ of heat?

6.54 x 103 kJ

Solution

The ΔH 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 C6H6.

Convert the mass of C6H6 to moles, and then calculate the amount of heat that will be released from 2 moles of C6H6 considering that burning 27.3 g C6H6 gives 1144 kJ of heat.

n(C6H6) = 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 C6H6 gives 1144 kJ heat, 2 mol C6H6 will give an unknown amount of heat:

0.350 mol C6H6 – 1144 kJ

2 mol C6H6 – X kJ

X = 6.54 x 103 kJ

Alternatively, we can do this by unit conversion method:

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

18.

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?

CH4(g) + 3Cl2(g) → CHCl3(l) + 3HCl(g), ΔH° = -334 kJ

713 kJ

Solution

The ΔH 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:

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

19.

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?

MgCO3(s) + 2HCl(aq)    MgCl2(aq) + H2O(l) + CO2(g), ΔH° = –112 kJ

14.6 kJ

Solution

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

n(MgCO3) = 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 MgCO3 or HCl could produce less product based on their moles and the reaction stoichiometry. Let’s do the calculations based on the number of MgCl2 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 MgCl2, 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:

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

20.

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

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g), ΔH°rxn = ?

ΔHf° for NH3(g) = –46.2 kJ/mol
ΔHf° for NO(g) = 90.4 kJ/mol
ΔHf° for H2O(g) = –241.8 kJ/mol

ΔH°rxn = -904.4 kJ

Solution

To calculate 𝚫H°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:

ΔH°rxn = ΣnpΔHof (products) – ΣnrΔH°f (reactants)

Where np and nr 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:

ΔH°rxn = [4 x ΔHof (NO) + 6 x ΔHof (H2O)] – [4 x ΔHof (NH3) + 5 x ΔHof (O2)]

ΔH°rxn = [4 x 90.4 kJ + 6 x (–241.8)] – [4 x (–46.2 kJ) + 5 x 0]

ΔH°rxn = -904.4 kJ

ΔH°rxn is negative which means it is an exothermic reaction releasing 904.40kJ of heat.

21.

Combustion of butane (C4H10) releases 5755 kJ of energy according to the following chemical equation.

2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l), ΔH°rxn = -5755 kJ

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

ΔHf° for CO2(g) = –393.5 kJ/mol
ΔHf° for H2O(l) = –285.8 kJ/mol

-126 kJ/mol

Solution

To calculate ΔHof (C4H10), we are going to use the equation for the heat of reaction based on the standard enthalpies of formation:

ΔH°rxn = ΣnpΔHof (products) – ΣnrΔH°f (reactants)

Where np and nr 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:

ΔH°rxn = [8 x ΔHof (CO2) + 10 x ΔHof (H2O)] – [2 x ΔHof (C4H10) + 13 x ΔHof (O2)] = -5755 kJ

ΔHof (C4H10) is the unknown that we need to solve for in this equation:

ΔH°rxn = [8 x (–393.5 kJ) + 10 x ΔHof (–285.8 kJ)] – [2 x ΔHof (C4H10) + 13 x 0] = -5755 kJ

-3148 kJ – 2858 kJ – 2 ΔHof (C4H10) = -5755 kJ

-3148 kJ – 2858 kJ + 5755 kJ = 2 ΔHof (C4H10)

ΔHof (C4H10) = -125.5 kJ/mol

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

22.

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) + 3O2(g) → 2ZnO(s) + 2SO2(g)

ΔHf° for SO2(g) = -296.8 kJ/mol
ΔHf° for ZnS(s) = -206.0 kJ/mol
ΔHf° for ZnO(s) = -350.5 kJ/mol

-883 kJ

Solution

To calculate 𝚫H°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:

ΔH°rxn = ΣnpΔHof (products) – ΣnrΔH°f (reactants)

Where np and nr 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:

ΔH°rxn = [2 x ΔHof (ZnO) + 2 x ΔHof (SO2)] – [2 x ΔHof (ZnS) + 3 x ΔHof (O2)]

ΔH°rxn = [2 x (-350.5 kJ) + 2 x (-296.8 kJ)] – [2 x (-206.0 kJ) + 3 x 0]

ΔH°rxn = -701 kJ – 593.6 kJ + 412 kJ = -883 kJ

ΔH°rxn is negative which means it is an exothermic reaction releasing 883 kJ of heat.

23.

Using the standard heats of formation given below, calculate the heat of reaction between barium carbonate (BaCO3) and sulfuric acid (H2SO4).

BaCO3(s) +  H2SO4(aq) → BaSO4(s) + CO2(g) + H2O(l)

ΔHf° for BaSO4(s) = -1473.2 kJ/mol
ΔHf° for CO2(g) = –393.5 kJ/mol
ΔHf° for H2O(l) = –285.8 kJ/mol
ΔHf° for BaCO3(s) = -1213.0 kJ/mol
ΔHf° for 3H2SO4(aq) = -814.0kJ/mol

-126 kJ

Solution

To calculate 𝚫H°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:

ΔH°rxn = ΣnpΔHof (products) – ΣnrΔH°f (reactants)

Where np and nr 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:

ΔH°rxn = [1 x ΔHof (BaSO4(s)) + 1 x ΔHof (CO2(g)) + 1 x ΔHof (H2O(l))] – [1 x ΔHof (BaCO3(s)) + 1 x ΔHof (H2SO4(aq))]

ΔH°rxn = [-1473.2 kJ –393.5 kJ –285.8 kJ] – [-1213.0 kJ -814.0kJ]

ΔH°rxn = -2,152.5 kJ + 2,027 kJ = -125.5 kj

ΔH°rxn is negative which means it is an exothermic reaction releasing 126 kJ of heat.

24.

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

2NO(g) + Cl2(g) ⇆ 2NOCl(g)

What is the enthalpy change for the following reaction?

2NOCl(g) ⇆ 2NO(g) + Cl2(g)

+76 kJ

Solution

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

25.

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

2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)

What is the enthalpy change for the following reaction?

H2S(g) + 1.5O2(g) → SO2(g) + H2O(g)

-518 kJ

Solution

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

26.

Consider the thermochemical equation for the combustion of nitromethane (CH3NO2):

2CH3NO2(l) + 3/2O2(g) → 2CO2(g) + 3H2O(l) + N2(g), ΔH °rxn = -1418 kJ

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

4CH3NO2(l) + 3O2(g) → 4CO2(g) + 6H2O(l) + N2(g), ΔH °rxn = ?

-2836 kJ

Solution

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

27.

Using the Hess’s law, calculate ΔHo for the reaction:

2NO2 + Cl2(g) → 2NOCl(g) + O2(g), ΔHo = ?

Use the following reactions and given ΔH’s:

1) 2NO(g) + Cl2(g) → 2NOCl(g), ΔHo = -76 kJ

2) 2NO(g) + O2(g) → 2NO2(g), ΔHo = -114 kJ

38 kJ

Solution

NO2 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 ΔHo when reversing a reaction!

2a) 2NO2(g) → 2NO(g) + O2(g), ΔHo = +114 kJ

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

2NO(g) + Cl2(g) → 2NOCl(g), ΔHo = -76 kJ
+
2NO2(g) → 2NO(g) + O2(g), ΔHo = +114 kJ

2NO(g) + Cl2(g) + 2NO2(g) → 2NOCl(g) + 2NO(g) + O2(g)

2NO2(g) + Cl2(g) → 2NOCl(g) + O2(g)

ΔHo = -76 kJ +114 kJ = 38 kJ

In summary, we can write:

2NO2 + Cl2(g) → 2NOCl(g) + O2(g), ΔHo = 38 kJ

28.

Calculate the enthalpy for the oxidation of CO to CO2 using the enthalpy of reaction for the combustion of C to CO (ΔH = -221.0 kJ) and the enthalpy for the combustion of C to CO2 (ΔH = -393.5 kJ).

2CO(g) + O2(g)  →  2CO2(g)  ΔH = ?

1) C(s) + O2(g)  →  CO2(g)  ΔH = -393.5 kJ

2) 2C(s) + O2(g)  →  2CO(g)  ΔH = -221.0 kJ

-566 kJ

Solution

The target reaction contains 2 moles of CO2 as a product, so we need to multiply equation 1 by two and do the same for the value of ΔH:

1a) 2C(s) + 2O2(g) → 2CO2(g), ΔH = -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) + O2(g), ΔH = 221.0 kJ

And now we can add the two equations:

1a) 2C(s) + 2O2(g) → 2CO2(g), ΔH = -787 kJ
+
2a) 2CO(g) → 2C(s) + O2(g), ΔH = 221.0 kJ

2C(s) + 2O2(g) + 2CO(g) → 2CO2(g) + 2C(s) + O2(g)

2CO(g) + O2(g) → 2CO2(g)

ΔH = -787 kJ + 221.0 kJ = -566 kJ

29.

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

2S(s) + 3O2(g)  →  2SO3(g ΔH = ?

1) S(s) + O2(g)  →  SO2(g)  ΔH = -297 kJ

2) 2SO3(g)  → 2SO2(g) + O2(g)  ΔH = 198 kJ

-792 kJ

Solution

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 ΔH:

1a) 2S(s) + 2O2(g)  →  2SO2(g)  ΔH = -594 kJ

The second equation needs to be reversed since in the target reaction, SO3 is a product while here it is a reactant:

2a) 2SO2(g) + O2(g) → 2SO3(g)  ΔH = -198 kJ

And now we can add the two equations:

1a) 2S(s) + 2O2(g)  →  2SO2(g)  ΔH = -594 kJ
+
2a) 2SO2(g) + O2(g) → 2SO3(g)  ΔH = -198 kJ

2S(s) + 2O2(g) + 2SO2(g) + O2(g) → 2SO2(g) + 2SO3(g)

2S(s) + 3O2(g) → 2SO3(g)

ΔH = -594 kJ + (-198 kJ) = -792 kJ

30.

Using the Hess’s law, calculate ΔHo for the combustion reaction of butene:

C4H8(g) + 6O2(g) → 4CO2(g) + 4H2O(l), ΔHo = ?

Use the following reactions and given ΔH’s:

1) 2H2(g) + O2(g) → 2H2O(g), ΔHo = -571 kJ

2) C4H8(g) + H2(g) → C4H10(g), ΔHo = -126 kJ

3) 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l), ΔHo = -5754 kJ

ΔHo = -2718 kJ

Solution

Because C4H8 appears as a reactant in the target reaction, we are not going to reverse equation 2. However, to get rid of C4H10, we need to multiply equation 2 by two since there are two moles of C4H10 in equation 3. Let’s label the new equation as 2a:

2a) 2C4H8(g) + 2H2(g) → 2C4H10(g), ΔHo = -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 C4H8 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) 2H2O(g) → 2H2(g) + O2(g), ΔHo = 571 kJ

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

1a) 2H2O(g) → 2H2(g) + O2(g), ΔHo = 571 kJ
+
2a) 2C4H8(g) + 2H2(g) → 2C4H10(g), ΔHo = -252 kJ
+
3) 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l), ΔHo = -5754 kJ

2H2O(g) + 2C4H8(g) + 2H2(g) + 2C4H10(g) + 13O2(g) → 2H2(g) + O2(g) + 2C4H10(g) + 8CO2(g) + 10H2O(l)

2C4H8(g) + 12O2(g) → 8CO2(g) + 8H2O(l)

ΔHo = 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:

2C4H8(g) + 12O2(g) → 8CO2(g) + 8H2O(l), ΔHo = -2718 kJ

31.

Using the Hess’s law and the enthalpies of the three combustion reactions below, calculate the enthalpy of the reaction producing methanol (CH3OH) from carbon monoxide and hydrogen gas.

CO(g) + 2H2(g) → CH3OH(g), ΔH = ?

1) 2CO(g) + O2(g) → 2CO2(g), ΔH = -566 kJ

2) 2H2(g) + O2(g) → 2H2O(g), ΔHo = -571 kJ

3) 2CH3OH(g) + 3O2(g) → 2CO2(g) + 4H2O(g), ΔHo = −1430 kJ

-139 kJ

Solution

The first two equations seem to be in the correct directions since they contain CO and H2 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) 2CO2(g) + 4H2O(g) → 2CH3OH(g) + 3O2(g), ΔHo = +1430 kJ

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

1) 2CO(g) + O2(g) → 2CO2(g), ΔH = -566 kJ
+
2) 2H2(g) + O2(g) → 2H2O(g), ΔHo = -571 kJ
+
3a) 2CO2(g) + 4H2O(g) → 2CH3OH(g) + 3O2(g), ΔHo = +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) 4H2(g) + 2O2(g) → 4H2O(g), ΔHo = -1141 kJ

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

1) 2CO(g) + O2(g) → 2CO2(g), ΔH = -566 kJ
+
2a) 4H2(g) + 2O2(g) → 4H2O(g), ΔHo = -1141 kJ
+
3a) 2CO2(g) + 4H2O(g) → 2CH3OH(g) + 3O2(g), ΔHo = +1430 kJ
__________________________________________________

2CO) + O2 + 4H2 + 2O2 + 2CO2 + 4H2O → 2CO2 + 4H2O + 2CH3OH + 3O2

Cancel the same molecules on both side of the equation:

2CO + O2 + 4H2 + 2O2 + 2CO2 + 4H2O → 2CO2 + 4H2O + 2CH3OH + 3O2

ΔHo = -566 kJ + (-1141 kJ) + 1430 kJ/mol = -277 kJ

So, we have:

2CO(g) + 4H2(g) → 2CH3OH(g), ΔHo = -277 kJ

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

CO(g) + 2H2(g) → CH3OH(g), ΔHo = -139 kJ

32.

Using the Hess’s law and the enthalpies of the given reactions, calculate the enthalpy of the combustion reaction of CH3Cl:

CH3Cl(g) + O2(g) → CO(g) + HCl(g) + H2O(l), ΔH = ?

1) CO(g) + 2H2(g) → CH3OH(g), ΔHo = -139 kJ

2) 2H2(g) + O2(g) → 2H2O(g), ΔHo = -571 kJ

3) CH3OH(g) + HCl(aq) → CH3Cl(g) + H2O(l), ΔHo = -28 kJ

-404 kJ

Solution

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) CH3OH(g) → CO(g) + 2H2(g), ΔHo = +139 kJ

The second reaction is good as it is. However, the third reaction needs to be reversed as well in order to make CH3Cl as a reactant as it is in the target equation:

3a) CH3Cl(g) + H2O(g) → CH3OH(g) + HCl(aq), ΔHo = +28 kJ

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

1a) CH3OH(g) → CO(g) + 2H2(g), ΔHo = +139 kJ
+
2) 2H2(g) + O2(g) → 2H2O(g), ΔHo = -571 kJ
+
3a) CH3Cl(g) + H2O(l) → CH3OH(g) + HCl(aq), ΔHo = +28 kJ
____________________________________________

CH3OH(g) + 2H2(g) + O2(g) + CH3Cl(g) + 2H2O(l) → CO(g) + 2H2(g) + 2H2O(g) + CH3OH(g) + HCl(aq)

Cancel the same molecules on both side of the equation:

CH3OH(g) + 2H2(g) + O2(g) + CH3Cl(g) + H2O(l) → CO(g) + 2H2(g) + 2H2O(g) + CH3OH(g) + HCl(aq)

CH3Cl(g) + O2(g) → CO(g) + HCl(g) + H2O(g)

ΔHo = +139 kJ + (-571 kJ) + 28 kJ = -404 kJ

33.

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

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g), ΔH = ?

1) N2(g) + 3H2(g) → 2NH3(g), ΔH = -92 kJ

2) 2H2(g) + O2(g) → 2H2O(g), ΔH = -484 kJ

3) N2(g) + O2(g) → 2NO(g), ΔH = 181 kJ

-906 kJ

Solution

The first reaction needs to be reversed because the NH3 is a product, while in the target reaction it is a reactant. Let’s reverse it and label it as 1a:

1a) 2NH3(g) → N2(g) + 3H2(g), ΔH = +92 kJ

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

1a) 2NH3(g) → N2(g) + 3H2(g), ΔH = +92 kJ

2) 2H2(g) + O2(g) → 2H2O(g), ΔH = -484 kJ

3) N2(g) + O2(g) → 2NO(g), ΔH = 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) 4NH3(g) → 2N2(g) + 6H2(g), ΔH = +184 kJ

2a) 6H2(g) + 3O2(g) → 6H2O(g), ΔH = -1452 kJ

And now let’s put these three equations together:

1b) 4NH3(g) → 2N2(g) + 6H2(g), ΔH = +184 kJ

2a) 6H2(g) + 3O2(g) → 6H2O(g), ΔH = -1452 kJ

3) N2(g) + O2(g) → 2NO(g), ΔH = 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) 2N2(g) + 2O2(g) → 4NO(g), ΔH = 362 kJ

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

1b) 4NH3(g) → 2N2(g) + 6H2(g), ΔH = +184 kJ
+
2a) 6H2(g) + 3O2(g) → 6H2O(g), ΔH = -1452 kJ
+
3a) 2N2(g) + 2O2(g) → 4NO(g), ΔH = 362 kJ
____________________________________________

4NH3(g) + 6H2(g) + 3O2(g) + 2N2(g) + 2O2(g) → 2N2(g) + 6H2(g) + 6H2O(g) + 2NO(g)

4NH3(g) + 5O2(g) → 6H2O(g) + 2NO(g)

ΔH = +184 kJ + (-1452 kJ) + 362 kJ = -906 kJ

34.

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) + Cl2(g) + 2H2O(g), ΔH = ?

1) CuO(s) + H2(g) → Cu(s) + H2O(g), ΔH = -85 kJ

2) 2Cu(s) + Cl2(g) → 2CuCl(s), ΔH = -274 kJ

3) H2(g) + Cl2(g) → 2HCl(g), ΔH = -184 kJ

-76 kJ

Solution

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) + 2H2(g) → 2Cu(s) + 2H2O(g), ΔH = -170 kJ

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

There are 2H2(g) in equation 1a, and this means that we need to reverse equation 3 and multiply it by two so that we can have 2H2(g) as a product and cancel them with the ones in equation 1a:

3a) 4HCl(g) → 2H2(g) + 2Cl2(g), ΔH = 368 kJ

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

1a) 2CuO(s) + 2H2(g) → 2Cu(s) + 2H2O(g), ΔH = -170 kJ

2) 2Cu(s) + Cl2(g) → 2CuCl(s), ΔH = -274 kJ

3a) 4HCl(g) → 2H2(g) + 2Cl2(g), ΔH = 368 kJ

2CuO(s) + 2H2(g) + 2Cu(s) + Cl2(g) + 4HCl(g) → 2Cu(s) + 2H2O(g) + 2CuCl(s) + 2H2(g) + 2Cl2(g)

2CuO(s) + 4HCl(g) → 2CuCl(s) + 2H2O(g) + Cl2(g)

ΔH = -170 kJ + (-274 kJ) + 368 kJ = -76 kJ