The enthalpy change for a chemical reaction (Δ*H*rxn), also known as the **enthalpy of reaction **or **heat of reaction **is often given next to the chemical equation. These are called *thermochemical equations.*

For example, the enthalpy change for the combustion reaction of methane (CH_{4}), the principal component of natural gas, is given as:

CH_{4}(*g*) + 2O_{2}(*g*) → CO_{2}(*g*) + 2H_{2}O(*g*) Δ*H*_{rxn} = -2890.4 kJ

It is important to correctly **interpret the information** in this equation. First, we conclude that it is an **exothermic** reaction because of the **negative sign of Δ****H**.

* *In addition, and this is the focus of this article, it tells us that when 1 mol of CH_{4} reacts with 2 mol of O_{2} to form 1 mol of CO_{2} and 2 mol of H_{2}O, 2890.4 kJ of heat is released. So, the enthalpy change is part of the stoichiometric relationship of the equation just like the molar coefficients.

Therefore, if we are asked to determine the enthalpy change for a certain amount of any reactant or product of the reaction, we can write these relationships as molar conversion factors and do the necessary calculation.

**For example**,

How much heat will be released if 5 moles of CH_{4 }is reacted with an excess of oxygen?

CH_{4}(*g*) + 2O_{2}(*g*) → CO_{2}(*g*) + 2H_{2}O(*g*) Δ*H*_{rxn} = -2890.4 kJ

According to the equation, 2890.4 kJ of heat is released when 1 mol of CH_{4 }is reacted. Therefore, the conversion factor would be:

\[\;\frac{{2890.4\;{\rm{kJ}}}}{{{\rm{1}}\;{\rm{mol}}\,{\rm{C}}{{\rm{H}}_{\rm{4}}}}}\;\]

So, to determine the ΔH for 5 moles of CH_{4}, we can write:

* *

\[\Delta H\; = \;\;\frac{{2890.4\;{\rm{kJ}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\,{\rm{C}}{{\rm{H}}_{\rm{4}}}}}}}\; \times \;5\cancel{{{\rm{mol}}\;{\rm{C}}{{\rm{H}}_{\rm{4}}}}}\;{\rm{ = }}\;14,452\;{\rm{kJ}}\]

Keep in mind that, even though you are in the thermochemistry chapter, there might be questions needing to address some concepts covered earlier in the semester. Like any chemical equation, be sure to balance it when needed, consider the limiting reactant, percent yield etc.

If the amount of a reaction component is given in grams or other units, convert it to moles first.

**For example**,

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

First, convert all the quantities to moles.

\[{\rm{n}}\;{\rm{(MgC}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\,{\rm{12}}{\rm{.65 }}\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{84}}{\rm{.3}}\;\cancel{{\rm{g}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.150 mol}}\]

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

\[{\rm{M}}\;{\rm{ = }}\;\frac{{\rm{n}}}{{\rm{V}}}\,\, \Rightarrow \;\,{\rm{n}}\,{\rm{ = }}\;{\rm{MV}}\]

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 amount of heat based 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 the unit conversion method shown in for the first reaction:

** **

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

**Δ****H When Changing the Chemical Equation**

**Δ**

As we mentioned earlier, the Δ*H*_{rxn} value corresponds to the given moles/coefficients of reactants and products. So;

**1)** If the **equation is multiplied **by a factor, then ** Hrxn must be multiplied **by the same factor:

**For example**, the following equation indicates that when 2 mol of CH_{3}OH reacts with 3 mol of O_{2} to form 2 mol of CO_{2} and 4 mol of H_{2}O, 1453 kJ of heat is released.

2CH_{3}OH(*l*) + 3O_{2}(*g*) → 2CO_{2}(*g*) + 4H_{2}O(*l*) Δ*H*_{rxn} = -1453 kJ

Now, if we multiply the coefficients by two, the enthalpy change must be multiplied by two as well:

2CH_{3}OH(*l*) + 3O_{2}(*g*) → 2CO_{2}(*g*) + 4H_{2}O(*l*) Δ*H*_{rxn} = -1453 kJ

*x2*

4CH_{3}OH(*l*) + 6O_{2}(*g*) → 4CO_{2}(*g*) + 8H_{2}O(*l*) Δ*H*_{rxn} = -2906 kJ

**2) **Δ*H*_{rxn}** changes the sign **when a chemical **equation is reversed:**

** **

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

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

**Check Also**

- 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
- Hess’s Law and Enthalpy of Reaction
- Hess’s Law Practice Problems
- Standard Enthalpies of Formation
- Enthalpy of Reaction from Enthalpies of Formation
**Thermochemistry Practice Problems**

** **

#### Practice

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?

2SO_{2}(*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:

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