The variation of with for a reaction is plotted as a straight line. If the line makes an angle of wi — Chemical Equilibrium Chemistry Question
Question
The variation of $\log_{10} K$ with $1/T$ for a reaction is plotted as a straight line. If the line makes an angle of $45^\circ$ with the positive x-axis (i.e., slope is 1), what is the standard enthalpy change ($\Delta H^\circ$) for the reaction? (Use $2.303 \times R \approx 19.14 \text{ J mol}^{-1} \text{K}^{-1}$)
Answer: A
💡 Solution & Explanation
According to the integrated Van 't Hoff equation, the slope of $\log_{10} K$ vs $1/T$ is exactly $-\frac{\Delta H^\circ}{2.303 R}$. Given the slope is $\tan(45^\circ) = 1$, we have $-\frac{\Delta H^\circ}{2.303 R} = 1$. Solving for $\Delta H^\circ$ gives $-2.303 R = -19.14 \text{ J mol}^{-1}$.
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