The values of pressure equilibrium constant r | vedclass

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Question

(C) The Van't Hoff equation is given by: $\ln K_p = -\frac{\Delta H^\circ}{R} \left(\frac{1}{T}\right) + C$.Converting $\log_{10}$ to $\ln$: $\ln K_p

Vedclass · Accepted Answer

$230$

Vedclass · Answer

(C) The Van't Hoff equation is given by: $\ln K_p = -\frac{\Delta H^\circ}{R} \left(\frac{1}{T}ight) + C$.Converting $\log_{10}$ to $\ln$: $\ln K_p = 2.303 \log_{10} K_p$.Substituting this into the equation: $2.303 \log_{10} K_p = -\frac{\Delta H^\circ}{R} \left(\frac{1}{T}ight) + C$.Dividing by $2.303$: $\log_{10} K_p = -\frac{\Delta H^\circ}{2.303 R} \left(\frac{1}{T}ight) + C'$.This is in the form of a straight line $y = mx + c$, where the slope $m = -\frac{\Delta H^\circ}{2.303 R}$.Using the given data points $(0.05, 3.5)$ and $(0.06, 2.5)$:Slope $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2.5 - 3.5}{0.06 - 0.05} = \frac{-1.0}{0.01} = -100$.Equating the slope: $-100 = -\frac{\Delta H^\circ}{2.303 R}$.Therefore, $\frac{\Delta H^\circ}{R} = 100 	imes 2.303 = 230.3 \approx 230$.

$1/T \text{ (K}^{-1})$	$\log_{10} K_p$
$0.05$	$3.5$
$0.06$	$2.5$
$0.07$	$1.5$

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