The value of standard enthalpy, $$\triangle H^{\ominus}$$ (in kJ mol$$^{-1}$$) for the given reaction is ___.

The graph provided is a straight line of $$\ln\dfrac{p_{z}}{p^{\ominus}}$$ (ordinate) versus $$\dfrac{10^{4}}{T}$$ (abscissa).
Let the coordinates of two well-defined points on the solid line be $$\left(x_{1},y_{1}\right)=\left(\dfrac{10^{4}}{T_{1}},\ \ln\dfrac{p_{z1}}{p^{\ominus}}\right)$$ and $$\left(x_{2},y_{2}\right)=\left(\dfrac{10^{4}}{T_{2}},\ \ln\dfrac{p_{z2}}{p^{\ominus}}\right).$$

The slope of the straight line with these scaled axes is $$m'=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}=\dfrac{\Delta\left(\ln K\right)}{\Delta\left(\dfrac{10^{4}}{T}\right)}.$$ From the plotted data, the two points that lie exactly on grid intersections are observed to be:

$$\left(2.0,\; 1.0\right)\quad\text{and}\quad\left(3.0,\; -1.0\right).$$ Therefore $$m'=\dfrac{-1.0-1.0}{3.0-2.0}=\dfrac{-2.0}{1.0}=-2.0.$$

Remember that the Van’t Hoff relation requires the derivative with respect to $$\dfrac{1}{T}$$, not $$\dfrac{10^{4}}{T}$$. Because $$\dfrac{10^{4}}{T}=10000\left(\dfrac{1}{T}\right),$$ we have

$$\frac{d\!\left(\ln K\right)}{d\!\left(\dfrac{1}{T}\right)} =\frac{d\!\left(\ln K\right)}{d\!\left(\dfrac{10^{4}}{T}\right)} \times \frac{d\!\left(\dfrac{10^{4}}{T}\right)}{d\!\left(\dfrac{1}{T}\right)} =m' \times 10000.$$ Substituting $$m'=-2.0,$$

$$\frac{d\!\left(\ln K\right)}{d\!\left(\dfrac{1}{T}\right)}=-2.0 \times 10000=-2.00\times10^{4}.$$

Using the Van’t Hoff equation $$\frac{d(\ln K)}{d\left(\frac{1}{T}\right)}=-\frac{\Delta H^{\ominus}}{R},$$ we get

$$-\frac{\Delta H^{\ominus}}{R}=-2.00\times10^{4}.$$ Hence $$\Delta H^{\ominus}=2.00\times10^{4}\,R.$$

With $$R=8.314\ \text{J K}^{-1}\text{ mol}^{-1},$$

$$\Delta H^{\ominus}=2.00\times10^{4}\times8.314 =1.6628\times10^{5}\ \text{J mol}^{-1}.$$

Converting to kJ mol$$^{-1}:$$ $$\Delta H^{\ominus}=166.28\ \text{kJ mol}^{-1}.$$

Therefore, the standard enthalpy change for the reaction is $$\Delta H^{\ominus}\;=\;166.28\ \text{kJ mol}^{-1}$$.