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 $1/T$ for a certain reversible reaction yields a straight line with a slope exactly equal to $-2500\text{ K}$, what is the exact integer value of the ratio $\frac{\Delta H^\circ}{R}$?
Answer: 2500
💡 Solution & Explanation
By comparing the given equation to the slope-intercept form of a straight line $y = mx + c$, where $y = \ln K$ and $x = 1/T$, the slope $m$ is exactly $-\Delta H^\circ / R$. Since we are given that the slope is $-2500$, we can write $-\Delta H^\circ / R = -2500$. Multiplying both sides by $-1$ yields $\Delta H^\circ / R = 2500$.
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