The variation of the equilibrium constant with temperature is given by the equation . For a specific — Chemical Equilibrium Chemistry Question
Question
The variation of the equilibrium constant with temperature is given by the equation $\ln K = -\frac{\Delta H^\circ}{RT} + \frac{\Delta S^\circ}{R}$. For a specific reaction, the plot of $\ln K$ versus $1/T$ yields a straight line with a slope of $-2500\text{ K}$. What is the value of $\frac{\Delta H^\circ}{R}$ for this reaction?
Answer: 2500
💡 Solution & Explanation
Comparing the given equation $\ln K = -\frac{\Delta H^\circ}{R} \left(\frac{1}{T}\right) + \frac{\Delta S^\circ}{R}$ with the equation of a straight line $y = mx + c$, where $y = \ln K$ and $x = 1/T$, the slope $m$ is exactly $-\frac{\Delta H^\circ}{R}$. We are given that the slope is $-2500$. Therefore, $-\frac{\Delta H^\circ}{R} = -2500$, which simplifies to $\frac{\Delta H^\circ}{R} = 2500$.
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