The values of pressure equilibrium constant recorded at different...

According to the van 't Hoff equation:

$\log_{10} K_p = -\dfrac{\Delta H^\circ}{2.303 R} \left(\dfrac{1}{T}\right) + \dfrac{\Delta S^\circ}{2.303 R}$

This represents a linear equation $y = mx + c$, where $y = \log_{10} K_p$ and $x = \dfrac{1}{T}$. The slope $m$ of the line is given by:

$m = -\dfrac{\Delta H^\circ}{2.303 R}$

From the given data table, we can calculate the slope $m$ using any two points, for example $(0.05, 3.5)$ and $(0.06, 2.5)$:

$m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{2.5 - 3.5}{0.06 - 0.05} = \dfrac{-1.0}{0.01} = -100$

Equating the calculated slope to the theoretical slope:

$-\dfrac{\Delta H^\circ}{2.303 R} = -100$

$\dfrac{\Delta H^\circ}{R} = 100 \times 2.303 = 230.3$

The magnitude of $\dfrac{\Delta H^\circ}{R}$ is $230.3$. Rounding to the nearest integer, we get $230$.

Answer: $230$