Question

Calculate Delta H in KJ for the following reactions using heats of formation:

Answer 1

Answer:

(a)

$\Delta H\textdegree = -2856.8\;\text{kJ}$ per mole reaction.

(b)

$\Delta H\textdegree = -22.3\;\text{kJ}$ per mole reaction.

Explanation:

What is the standard enthalpy of formation $\Delta H_f\textdegree{}$ of a substance? $\Delta H_f\textdegree{}$ the enthalpy change when one mole of the substance is formed from the most stable allotrope of its elements under standard conditions.

Naturally, $\Delta H_f\textdegree{} = 0$ for the most stable allotrope of each element under standard conditions. For example, oxygen $\text{O}_2$ (not ozone $\text{O}_3$) is the most stable allotrope of oxygen. Also, under STP $\text{O}_2$  is a gas. Forming $\text{O}_2\;(g)$ from itself does not involve any chemical or physical change. As a result, $\Delta H_f\textdegree{} = 0$ for $\text{O}_2\;(g)$.

Look up standard enthalpy of formation $\Delta H_f\textdegree{}$ data for the rest of the species. In case one or more values are not available from your school, here are the published ones. Note the state symbols of the compounds (water/steam $\text{H}_2\text{O}$ in particular) and the sign of the enthalpy changes.

• $\text{C}_2\text{H}_6\;(g)$: $-84.0\;\text{kJ}\cdot\text{mol}^{-1}$;
• $\text{CO}_2\;(g)$: $-393.5\;\text{kJ}\cdot\text{mol}^{-1}$;
• $\text{H}_2\text{O}\;{\bf (g)}$: $-241.8\;\text{kJ}\cdot\text{mol}^{-1}$;
• $\text{PbO}\;(s)$: $-217.9\;\text{kJ}\cdot\text{mol}^{-1}$;
• $\text{PbO}_2\;(s)$: $-276.6\;\text{kJ}\cdot\text{mol}^{-1}$;
• $\text{Pb}_3\text{O}_4\;(s)$: $-734.7\;\text{kJ}\cdot\text{mol}^{-1}$

How to calculate the enthalpy change of a reaction $\Delta H_\text{rxn}$ (or simply $\Delta H$ from enthalpies of formation?

• Multiply the enthalpy of formation of each product by its coefficient in the equation.
• Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Reactants}))$ to show that this value takes the coefficients into account.
• Multiply the enthalpy of formation of each reactant by its coefficient in the equation.
• Find the sum of these values. Label the sum $\Sigma (n\cdot \Delta_f(\text{Products}))$ to show that this value takes the coefficient into account.
• Change = Final - Initial. So is the case with enthalpy changes. $\Delta H_\text{rxn} = \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))$.

For the first reaction:

• $\Sigma (n\cdot \Delta_f(\text{Reactants})) = 4\times (-393.5) + 6\times (-241.8) = -3024.8\;\text{kJ}\cdot\text{mol}^{-1}$;
• $\Sigma (n\cdot \Delta_f(\text{Products})) = 2\times (-84.0) + 7\times 0 = -168.0\;\text{kJ}\cdot\text{mol}^{-1}$;
• $\begin{aligned}\Delta H_\text{rxn} &= \Sigma (n\cdot \Delta_f(\textbf{Products})) - \Sigma (n\cdot \Delta_f(\textbf{Reactants}))\\ &= (-3024.8\;\text{kJ}\cdot\text{mol}^{-1}) - (-168.0\;\text{kJ}\cdot\text{mol}^{-1})\\ &= -2856.8\;\text{kJ}\cdot\text{mol}^{-1} \end{aligned}$.

Try these steps for the second reaction:

$\Delta H_\text{rxn} = -22.3\;\text{kJ}\cdot\text{mol}^{-1}$.