The enthalpy change for a chemical reaction (ΔHrxn), 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 (CH₄), the principal component of natural gas, is given as:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) ΔH_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 ΔH.

 In addition, and this is the focus of this article, it tells us that when 1 mol of CH₄ reacts with 2 mol of O₂ to form 1 mol of CO₂ and 2 mol of H₂O, 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 CH₄ is reacted with an excess of oxygen?

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) ΔH_rxn = -2890.4 kJ

According to the equation, 2890.4 kJ of heat is released when 1 mol of CH₄ 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 CH₄, 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?

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

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

ΔH When Changing the Chemical Equation

As we mentioned earlier, the ΔH_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 CH₃OH reacts with 3 mol of O₂ to form 2 mol of CO₂ and 4 mol of H₂O, 1453 kJ of heat is released.

2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(l) ΔH_rxn = -1453 kJ

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

2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(l) ΔH_rxn = -1453 kJ

x2

4CH₃OH(l) + 6O₂(g) → 4CO₂(g) + 8H₂O(l) ΔH_rxn = -2906 kJ

2) ΔH_rxn changes the sign when a chemical equation is reversed:

CH₄(g) + 3Cl₂(g) → CHCl₃(l) + 3HCl(g) H° = –334 kJ

CHCl₃(l) + 3HCl(g) → CH₄(g) + 3Cl₂(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