The variation of the equilibrium constant with absolute temperature is given by . If the plot of ver — Chemical Equilibrium Chemistry Question
Question
The variation of the equilibrium constant with absolute temperature is given by $\ln K = -\frac{\Delta H^\circ}{RT} + \frac{\Delta S^\circ}{R}$. If the plot of $\ln K$ versus $1/T$ yields a straight line with a slope exactly equal to $-2500\text{ K}$, what is the value of the ratio $\Delta H^\circ / R$?
Answer: A
💡 Solution & Explanation
Comparing the equation to $y = mx + c$, where $y = \ln K$ and $x = 1/T$, the slope $m = -\Delta H^\circ / R$. Given the slope is $-2500$, we have $-\Delta H^\circ / R = -2500$, which implies $\Delta H^\circ / R = 2500$.
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