Question

using the given information, calculate the standard enthalpy of reaction (in kj) for the following reaction.
cuo(s) + h₂(g) → cu(s) + h₂o(l)
report your answer to 1 place past the decimal point. if your answer is negative, include the negative sign. if it is positive, do not include the sign.

Explanation:

Step1: Recall the formula for standard enthalpy of reaction

The formula for the standard enthalpy of reaction ($\Delta H_{rxn}^{\circ}$) is $\Delta H_{rxn}^{\circ}=\sum n\Delta H_{f}^{\circ}(\text{products})-\sum m\Delta H_{f}^{\circ}(\text{reactants})$, where $n$ and $m$ are the stoichiometric coefficients of products and reactants respectively.

Step2: Identify reactants and products with their $\Delta H_{f}^{\circ}$

For the reaction $\text{CuO}(s)+\text{H}_2(g)
ightarrow\text{Cu}(s)+\text{H}_2\text{O}(l)$:

• Reactants: $\text{CuO}(s)$ with $\Delta H_{f}^{\circ}=- 156.1\space kJ/mol$ and $\text{H}_2(g)$ with $\Delta H_{f}^{\circ}=0\space kJ/mol$ (stoichiometric coefficient for both is 1).
• Products: $\text{Cu}(s)$ with $\Delta H_{f}^{\circ}=0\space kJ/mol$ and $\text{H}_2\text{O}(l)$ with $\Delta H_{f}^{\circ}=-285.8\space kJ/mol$ (stoichiometric coefficient for both is 1).

Step3: Calculate $\sum n\Delta H_{f}^{\circ}(\text{products})$

$\sum n\Delta H_{f}^{\circ}(\text{products})=(1\times\Delta H_{f}^{\circ}(\text{Cu}(s)))+(1\times\Delta H_{f}^{\circ}(\text{H}_2\text{O}(l)))=0+(-285.8)= - 285.8\space kJ/mol$

Step4: Calculate $\sum m\Delta H_{f}^{\circ}(\text{reactants})$

$\sum m\Delta H_{f}^{\circ}(\text{reactants})=(1\times\Delta H_{f}^{\circ}(\text{CuO}(s)))+(1\times\Delta H_{f}^{\circ}(\text{H}_2(g)))=-156.1 + 0=-156.1\space kJ/mol$

Step5: Calculate $\Delta H_{rxn}^{\circ}$

$\Delta H_{rxn}^{\circ}=\sum n\Delta H_{f}^{\circ}(\text{products})-\sum m\Delta H_{f}^{\circ}(\text{reactants})=(-285.8)-(-156.1)=-285.8 + 156.1=-129.7\space kJ/mol$

Answer:

-129.7