JEE Mains · Chemistry · STD 12 - 3. Chemical kinetics
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).
- A \(\frac{2.303R}{\Delta H^{\circ}}, \frac{2.303R}{\Delta S^{\circ}}\)
- B \(\frac{\Delta S^{\circ}}{2.303R}, -\frac{\Delta H^{\circ}}{2.303R}\)
- C -\(\frac{\Delta S^{\circ}R}{2.303}, \frac{\Delta H^{\circ}R}{2.303}\)
- D \(-\frac{\Delta H ^{\circ}}{2.303 R }, \frac{\Delta S ^{\circ}}{2.303 R }\)
Answer & Solution
Correct Answer
(B) \(\frac{\Delta S^{\circ}}{2.303R}, -\frac{\Delta H^{\circ}}{2.303R}\)
Step-by-step Solution
Detailed explanation
\(log_{10} K = -\frac{\Delta H^{\circ}}{2.303RT} + \frac{\Delta S^{\circ}}{2.303 R}\) \(y-intercept = \frac{\Delta S^{\circ}}{2.303 R}\) \(Slope = -\frac{\Delta H^{\circ}}{2.303 R}\)
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