Calculate Reaction Enthalpy from Equilibrium Constants at Different Temperatures

To calculate the magnitude of $\frac{\Delta H^\circ}{R}$, we use the van 't Hoff equation expressed in terms of base-10 logarithm:

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

where $C$ is a constant. This equation represents a linear relationship of the form $y = mx + c$, where:

• $y = \log_{10} K_p$
• $x = \frac{1}{T}$
• $\text{Slope } (m) = -\frac{\Delta H^\circ}{2.303 R}$

Using two data points from the given table:

1. $\left(\left(\frac{1}{T}\right)_1, (\log_{10} K_p)_1\right) = (0.05, 3.5)$
2. $\left(\left(\frac{1}{T}\right)_2, (\log_{10} K_p)_2\right) = (0.06, 2.5)$

The slope ($m$) of the line is: $m = \frac{(\log_{10} K_p)_2 - (\log_{10} K_p)_1}{\left(\frac{1}{T}\right)_2 - \left(\frac{1}{T}\right)_1} = \frac{2.5 - 3.5}{0.06 - 0.05} = \frac{-1.0}{0.01} = -100$

Now, equating the slope to the expression involving $\Delta H^\circ$: $-\frac{\Delta H^\circ}{2.303 R} = -100$

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

Taking the magnitude and rounding to the nearest integer: $\left|\frac{\Delta H^\circ}{R}\right| = 230$