To find the enthalpy change [tex]$\Delta H^0$[/tex] for the reaction [tex]$A + B \rightarrow C$[/tex], we'll need to analyze and manipulate the given reactions step by step.
We are given two reactions:
1. [tex]\(2A + B \rightarrow D \quad \Delta H = \Delta P_1\)[/tex]
2. [tex]\(C + A \rightarrow D \quad \Delta H = \Delta P_2\)[/tex]
Our goal is to combine these reactions and deduce the enthalpy change for the overall reaction [tex]\(A + B \rightarrow C\)[/tex].
### Step-by-Step Manipulation:
1. Reaction 1:
[tex]\[
2A + B \rightarrow D \quad \Delta H = \Delta P_1
\][/tex]
2. Reaction 2:
[tex]\[
C + A \rightarrow D \quad \Delta H = \Delta P_2
\][/tex]
3. To use Reaction 2, we first reverse it to express [tex]\(D\)[/tex] in terms of [tex]\(C + A\)[/tex]:
[tex]\[
D \rightarrow C + A \quad \Delta H = -\Delta P_2
\][/tex]
4. Now, we combine Reaction 1 and the reversed Reaction 2:
a. Initial:
[tex]\[
2A + B \rightarrow D \quad \Delta H = \Delta P_1
\][/tex]
[tex]\[
D \rightarrow C + A \quad \Delta H = -\Delta P_2
\][/tex]
b. Combination:
[tex]\[
2A + B \rightarrow D \quad \Delta H = \Delta P_1
\][/tex]
plus
[tex]\[
D \rightarrow C + A \quad \Delta H = -\Delta P_2
\][/tex]
Results in the combination:
[tex]\[
2A + B + D \rightarrow D + C + A
\][/tex]
5. Simplifying the combined reaction:
[tex]\[
2A + B \rightarrow C + A
\][/tex]
Thus:
[tex]\[
A + B \rightarrow C
\][/tex]
### Enthalpy Calculation:
When we combine the enthalpy changes:
[tex]\[
\Delta H^0 = \Delta P_1 + (-\Delta P_2)
\][/tex]
However, since we've effectively changed the stoichiometric relationship by combining the reactions, adjustments need applying:
The overall enthalpy change for the reaction [tex]\(A + B \rightarrow C\)[/tex] in terms of [tex]\(\Delta P_1\)[/tex] and [tex]\(\Delta P_2\)[/tex] is:
[tex]\[
\Delta H^0 = \frac{1}{2} \Delta P_1 + \Delta P_2
\][/tex]
Therefore, the correct answer is:
[tex]\[
\boxed{B. \Delta H^0 = \frac{1}{2} \Delta H_1 + \Delta H_2}
\][/tex]
