Question

you need to calculate the enthalpy change ($\delta h^{\circ}$) of the reaction a + b → c. how can the enthalpies given for the reaction steps below be combined to give the overall change in enthalpy? 2a + b→d $\delta h^{\circ}_1$ c + a → d $\delta h^{\circ}_2$ a. $\delta h^{\circ}=\delta h^{\circ}_1 - \delta h^{\circ}_2$ b. $\delta h^{\circ}=\delta h^{\circ}_2 - \delta h^{\circ}_1$ c. $\delta h^{\circ}=\delta h^{\circ}_1+\delta h^{\circ}_2$ d. $\delta h^{\circ}=\frac{1}{2}\delta h^{\circ}_1+\delta h^{\circ}_2$

Explanation:

Step1: Reverse the second reaction

When we reverse the reaction $C + A
ightarrow D$ (with $\Delta H^0_2$), we get $D
ightarrow C + A$ and the enthalpy change becomes $-\Delta H^0_2$.

Step2: Add the modified second - reaction to the first reaction

The first reaction is $2A + B
ightarrow D$ with $\Delta H^0_1$. Adding $D
ightarrow C + A$ (with enthalpy $-\Delta H^0_2$) to it:
$(2A + B
ightarrow D)+(D
ightarrow C + A)$ gives $A + B
ightarrow C$.
The overall enthalpy change $\Delta H^0$ for $A + B
ightarrow C$ is $\Delta H^0=\Delta H^0_1-\Delta H^0_2$.

Answer:

A. $\Delta H^0=\Delta H^0_1-\Delta H^0_2$