The plot of log_10 ⁡ K vs 1 T gives a straight line. The intercept and slope respectively are (where K is equilibrium constant).

Question

Accepted Answer

Correct answer: A. (a) The relationship between the standard Gibbs free energy change and the equilibrium constant is given by G^o=- RTln ⁡K. We also know that G^o= H^o-T S^o. Equating the two expressions: - RTln ⁡K= H^o-T S^o. Dividing by - RT, we get: ln ⁡K=- H^o RT+ S^oR. To convert natural log to base 10, use ln ⁡K=2.303 log_10 ⁡K : 2.303 log_10 ⁡K=- H^o RT+ S^oR. Dividing by 2.303 : log_10 ⁡K=- H^o2.303 RT+ S^o2.303R. Rearranging in the form of a straight-line equation y= mx+c where y= log_10 ⁡K and x= 1T : log_10 ⁡K=(- H^o2.303R) 1T+ S^o2.303R. Comparing the terms: Intercept (c )= S^o2.303R Slope (m )=- H^o2.303R.

$The plot of log_{10} K vs \frac{1}{T} gives a straight line. The intercept and slope respectively are (where K is equilibrium constant).$

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