1. Given relation

For the equilibrium $X(s) \rightleftharpoons Y(s) + Z(g),$ we are given $K = \frac{p_Z}{p^\theta}$ and $\frac{d(\ln K)}{d(1/T)} = -\frac{\Delta H^\circ}{R}.$

So, if we know the slope of the graph of $\ln(p_Z/p^\theta)$ versus $1/T$, we can find $\Delta H^\circ$.

2. Graph variable transformation

The graph is plotted against $x = \frac{10^4}{T}.$

Hence, $\frac{1}{T} = \frac{x}{10^4}.$

Therefore, $\frac{d(\ln K)}{d(1/T)} = \frac{d(\ln K)}{d(x/10^4)} = 10^4\,\frac{d(\ln K)}{dx}.$

So if the slope of the given graph of $\ln K$ vs $10^4/T$ is $m$, then $\frac{d(\ln K)}{d(1/T)} = 10^4 m.$

Thus, $-\frac{\Delta H^\circ}{R} = 10^4 m$ $\Rightarrow \Delta H^\circ = -R\,(10^4 m).$

3. Read slope from the graph

From the straight line (solid line), the slope is negative and equals approximately $m = -2.0.$

So, $\Delta H^\circ = -8.314 \times 10^4 \times (-2.0)\;\text{J mol}^{-1}.$

$\Delta H^\circ = 1.6628 \times 10^5\;\text{J mol}^{-1}.$

Converting to kJ mol$^{-1}$, $\Delta H^\circ = 166.28\;\text{kJ mol}^{-1}.$

4. Final answer

$\boxed{166.28}$

5. Comparison with stored answer

Stored correct answer = 166.28

My derived answer matches the stored answer.