ThermodynamicsHard
Question
The plot of $\log_{10}\text{ }K$ vs $\frac{1}{\text{ }T}$ gives a straight line. The intercept and slope respectively are (where K is equilibrium constant).
Options
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.$- \frac{\Delta H^{\circ}}{2.303R},\frac{\Delta S^{\circ}}{2.303R}$
Solution
$\log_{10}\text{ }K = - \frac{\Delta H^{o}}{2.303RT} + \frac{\Delta S^{o}}{2.303R}$
y -intercept $= \frac{\Delta S^{\circ}}{2.303R}$
Slope $= - \frac{\Delta H^{o}}{2.303R}$
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