To calculate the enthalpy change for the reaction:
[tex]\[
2 \text{ Hg} + \text{ Cl}_2 \rightarrow \text{ Hg}_2\text{Cl}_2
\][/tex]
we will use the given thermochemical equations:
1) [tex]\[
\text{HgCl}_2 \rightarrow \text{Hg} + \text{Cl}_2, \Delta H_1 = +224 \, \text{kJ}
\][/tex]
2) [tex]\[
\text{Hg} + \text{HgCl}_2 \rightarrow \text{Hg}_2\text{Cl}_2, \Delta H_2 = -41 \, \text{kJ}
\][/tex]
### Step-by-Step Solution:
1. Reverse the first reaction to match part of the overall reaction:
Original Reaction: [tex]\(\text{HgCl}_2 \rightarrow \text{Hg} + \text{Cl}_2\)[/tex]
Reversed Reaction: [tex]\(\text{Hg} + \text{Cl}_2 \rightarrow \text{HgCl}_2\)[/tex]
When reversing a reaction, the sign of [tex]\(\Delta H\)[/tex] is also reversed:
[tex]\[
\Delta H_{\text{reversed}} = - \Delta H_1 = -224 \, \text{kJ}
\][/tex]
2. Combine the reversed first reaction with the second reaction to obtain the overall reaction:
Reversed Reaction: [tex]\(\text{Hg} + \text{Cl}_2 \rightarrow \text{HgCl}_2\)[/tex]
Second Reaction: [tex]\(\text{Hg} + \text{HgCl}_2 \rightarrow \text{Hg}_2\text{Cl}_2\)[/tex]
Adding these two reactions together:
- LHS (left-hand side): [tex]\(\text{Hg} + \text{Cl}_2 + \text{Hg} + \text{HgCl}_2\)[/tex]
- RHS (right-hand side): [tex]\(\text{HgCl}_2 + \text{Hg}_2\text{Cl}_2\)[/tex]
[tex]\[
\text{Hg} + \text{Cl}_2 + \text{Hg} + \text{HgCl}_2 \rightarrow \text{HgCl}_2 + \text{Hg}_2\text{Cl}_2
\][/tex]
Simplifying:
[tex]\[
2 \text{Hg} + \text{Cl}_2 \rightarrow \text{Hg}_2\text{Cl}_2
\][/tex]
3. Combine the enthalpy changes for the reactions to find the total enthalpy change:
[tex]\[
\Delta H_{\text{total}} = \Delta H_{\text{reversed}} + \Delta H_2
\][/tex]
Substituting the known values:
[tex]\[
\Delta H_{\text{total}} = -224 \, \text{kJ} + (-41 \, \text{kJ}) = -265 \, \text{kJ}
\][/tex]
### Conclusion:
The enthalpy change for the reaction
[tex]\[
2 \text{ Hg} + \text{ Cl}_2 \rightarrow \text{ Hg}_2\text{Cl}_2
\][/tex]
is
[tex]\[
\Delta H_{\text{reaction}} = -265 \, \text{kJ}
\][/tex]
