The value of $$\triangle S^{\ominus}$$ (in JK$$^{-1}$$ mol$$^{-1}$$) for the given reaction, at 1000 K is ___.

For the equilibrium
$$X(s)\;\rightleftharpoons\;Y(s)+Z(g)$$ we have $$K=\dfrac{p_Z}{p^{\ominus}}$$ and the plot given is $$\ln K$$ versus $$\dfrac{10^{4}}{T}\,$$.

Step 1: Obtain $$\Delta H^{\ominus}$$ from the slope
Choose two clearly readable points on the straight line (solid line on the graph):
$$A:\;\Bigl(\dfrac{10^{4}}{T_1}=8,\;\ln K_1 =11\Bigr),\qquad B:\;\Bigl(\dfrac{10^{4}}{T_2}=12,\;\ln K_2 =8\Bigr)$$

Slope of the line
$$m=\dfrac{\ln K_2-\ln K_1}{\dfrac{10^{4}}{T_2}-\dfrac{10^{4}}{T_1}} =\dfrac{8-11}{12-8} =-\dfrac{3}{4} =-0.75$$

The two abscissae differ by a factor of $$10^{4}$$ from $$1/T$$, hence
$$\frac{d(\ln K)}{d\!\left(\frac{1}{T}\right)}=m\times10^{4} =(-0.75)\times10^{4} =-7.5\times10^{3}$$

Using the relation $$\dfrac{d(\ln K)}{d\!\left(\dfrac{1}{T}\right)}=-\dfrac{\Delta H^{\ominus}}{R}$$,
$$-\dfrac{\Delta H^{\ominus}}{R}=-7.5\times10^{3} \;\Longrightarrow\; \Delta H^{\ominus}=7.5\times10^{3}R$$

With $$R=8.314\ \text{J K}^{-1}\text{ mol}^{-1}$$,
$$\Delta H^{\ominus}=7.5\times10^{3}\times8.314 \approx6.24\times10^{4}\ \text{J mol}^{-1} =62.4\ \text{kJ mol}^{-1}$$

Step 2: Read $$\ln K$$ at $$T=1000\ \text{K}$$
Equation of the line:
$$\ln K-11=m\!\left(\dfrac{10^{4}}{T}-8\right)$$
At $$T=1000\ \text{K}$$, $$\dfrac{10^{4}}{T}=10$$, hence
$$\ln K=11+(-0.75)(10-8)=11-1.5=9.5$$

Step 3: Calculate $$\Delta S^{\ominus}$$
The thermodynamic identity
$$\Delta G^{\ominus}=\Delta H^{\ominus}-T\Delta S^{\ominus} \quad\text{with}\quad \Delta G^{\ominus}=-RT\ln K$$
gives
$$\Delta S^{\ominus}=\dfrac{\Delta H^{\ominus}}{T}+R\ln K$$

Compute each term at $$T=1000\ \text{K}$$:
$$\dfrac{\Delta H^{\ominus}}{T}=\dfrac{62.4\times10^{3}}{1000}=62.4\ \text{J K}^{-1}\text{ mol}^{-1}$$
$$R\ln K=8.314\times9.5\approx79.0\ \text{J K}^{-1}\text{ mol}^{-1}$$

Therefore
$$\Delta S^{\ominus}=62.4+79.0\approx141.4\ \text{J K}^{-1}\text{ mol}^{-1}$$

The required entropy change at $$1000\ \text{K}$$ is
$$\boxed{\;141\ \text{-}\ 142\ \text{J K}^{-1}\text{ mol}^{-1}\;}$$