Question

you need to calculate the enthalpy change ($\triangle h^0$) of the reaction a + 2b $\
ightarrow$ c. how can the enthalpies given for the reaction steps below be combined to give the overall change in enthalpy?
d + b $\
ightarrow$ a $\triangle h^0_1$
d + 3b $\
ightarrow$ c $\triangle h^0_2$
\bigcirc a. $\triangle h^0 = \triangle h^0_2 - \triangle h^0_1$
\bigcirc b. $\triangle h^0 = \triangle h^0_1 + \triangle h^0_2$
\bigcirc c. $\triangle h^0 = 2\triangle h^0_1 + \triangle h^0_2$
\bigcirc d. $\triangle h^0 = \triangle h^0_1 - \triangle h^0_2

Explanation:

Step1: Reverse the first reaction

The first reaction is \( D + B
ightarrow A \) with \( \Delta H^{\circ}_1 \). Reversing it gives \( A
ightarrow D + B \) and the enthalpy becomes \( -\Delta H^{\circ}_1 \) (since reversing a reaction changes the sign of \( \Delta H \)).

Step2: Add the reversed first reaction to the second reaction

The second reaction is \( D + 3B
ightarrow C \) with \( \Delta H^{\circ}_2 \). Adding the reversed first reaction (\( A
ightarrow D + B \), \( \Delta H = -\Delta H^{\circ}_1 \)) to the second reaction:
\( A + D + 3B
ightarrow D + B + C \)
Simplify by canceling \( D \) on both sides: \( A + 2B
ightarrow C \)
The total enthalpy change is \( \Delta H^{\circ} = \Delta H^{\circ}_2 + (-\Delta H^{\circ}_1) = \Delta H^{\circ}_2 - \Delta H^{\circ}_1 \)

Answer:

A. \( \Delta H^{\circ} = \Delta H^{\circ}_2 - \Delta H^{\circ}_1 \)