The enthalpy change for the reaction $$ \mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ is \(-891 \mathrm{~kJ}\) for the reaction as written. 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?

According to the balanced chemical equation for the reaction, $$ \mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ the overall enthalpy change of the reaction after the complete process is \(-891\) kJ. During the reaction, 2 moles of water are being formed. To find the quantity of heat released for each mole of water formed, simply divide the given enthalpy change by the number of moles of water produced in the reaction. $$\text{Heat released per mole of water} = \frac{-891\ \text{kJ}}{2\ \text{moles of }\mathrm{H_2O}}$$ Solve for the quantity of heat released per mole of water: $$\frac{-891\ \text{kJ}}{2\ \text{moles of }\mathrm{H_2O}} = -445.5\ \text{kJ/mole of }\mathrm{H_2O}$$ Therefore, \(445.5\ \text{kJ}\) of heat is released for each mole of water formed.

According to the balanced chemical equation for the reaction, $$ \mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ the overall enthalpy change for the reaction is -891 kJ. During the reaction, 2 moles of oxygen are being consumed. To find the quantity of heat released for each mole of oxygen reacted, divide the given enthalpy change by the number of moles of oxygen consumed in the reaction. $$\text{Heat released per mole of oxygen} = \frac{-891\ \text{kJ}}{2\ \text{moles of }\mathrm{O_2}}$$ Solve for the quantity of heat released per mole of oxygen: $$\frac{-891\ \text{kJ}}{2\ \text{moles of }\mathrm{O_2}} = -445.5\ \text{kJ/mole of }\mathrm{O_2}$$ Therefore, \(445.5\ \text{kJ}\) of heat is released for each mole of oxygen reacted.