The plot of log_{10} K vs frac{1}{T} gives a | vedclass

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(B) The relationship between equilibrium constant $K$ and temperature $T$ is given by the van't Hoff equation: $\ln K = -\frac{\Delta H^{\circ}}{RT} +

Vedclass · Accepted Answer

$\frac{\Delta S^{\circ}}{2.303R}, -\frac{\Delta H^{\circ}}{2.303R}$

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(B) The relationship between equilibrium constant $K$ and temperature $T$ is given by the van't Hoff equation: $\ln K = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R}$.Converting to base $10$ logarithm: $log_{10} K = -\frac{\Delta H^{\circ}}{2.303RT} + \frac{\Delta S^{\circ}}{2.303R}$.Comparing this with the linear equation $y = mx + c$,where $y = log_{10} K$ and $x = \frac{1}{T}$:The slope $m = -\frac{\Delta H^{\circ}}{2.303R}$.The $y$-intercept $c = \frac{\Delta S^{\circ}}{2.303R}$.Thus,the intercept and slope are $\frac{\Delta S^{\circ}}{2.303R}$ and $-\frac{\Delta H^{\circ}}{2.303R}$ respectively.

The plot of $log_{10} K$ vs $\frac{1}{T}$ gives a straight line. The intercept and slope respectively are (where $K$ is equilibrium constant).

Similar Questions

At $298 \, K$,for which of the following reactions will the value of $\frac{K_p}{K_c}$ be maximum and minimum,respectively?
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For which of the following reactions will $K_p = K_c$?

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