Question

For a certain equilibrium over the temperature range 500–700 K, the equilibrium constant obeys the equation

$$\log K_p = 30.1 - \frac{6.36\times10^3}{T} - 5.7\,\log T.$$

Find $\Delta H$ and $\Delta S$ at $T=600$ K. Then express $\Delta H$ in the form $A + B\,T$ and determine the values of $A$ and $B$.

Answer 1

$\Delta H = A + B\,T$ with
$A = 8.314\;(6.36\times10^3)\,(2.303) \approx 1.2185\times10^5\ \mathrm{J\,mol^{-1}}$,
$B = 5.7\,(8.314)\approx 47.4\ \mathrm{J\,K^{-1}\,mol^{-1}}$.
At $T=600$ K:
$\Delta H = 1.2185\times10^5 + (47.4)(600) \approx 1.5029\times10^5\ \mathrm{J\,mol^{-1}}$,
$\Delta S \approx 320.7\ \mathrm{J\,K^{-1}\,mol^{-1}}$.

Answer 2

Key relations

• $\ln K_p = \ln(10)\,\bigl(30.1 - \tfrac{6360}{T} - 5.7\,\log T\bigr)$
• $\displaystyle \Delta H = -R\,\frac{d(\ln K_p)}{d(1/T)}$
• $\Delta G = -RT\ln K_p = \Delta H - T\Delta S$

Step 1: Compute $\Delta H$ as a function of $T$

$$\frac{d(\ln K_p)}{d(1/T)} = \ln(10)\,\Bigl[6360 + 5.7\,T\Bigr],$$

so

$$\Delta H = R\,\ln(10)\,\bigl[6360 + 5.7\,T\bigr] = \underbrace{R\,(6360)\,\ln10}_{A} \;+\;\underbrace{R\,(5.7)}_{B}\,T.$$

Thus

$$A = 8.314\times6360\times2.303 \approx 1.2185\times10^5\ \mathrm{J/mol},\quad B = 5.7\times8.314 \approx 47.4\ \mathrm{J/K\,mol}.$$

Step 2: Evaluate $\Delta H$ at $T=600$ K

$$\Delta H = 1.2185\times10^5 + 47.4\times600 \approx 1.5029\times10^5\ \mathrm{J/mol}.$$

Step 3: Compute $\Delta S$
First, $\log K_p=30.1 - \tfrac{6360}{600}-5.7\log(600)=3.67$.
So $\ln K_p = 2.303\times3.67 \approx 8.45$. Then

$$\Delta G = -RT\ln K_p = -8.314\times600\times8.45 \approx -4.2145\times10^4\ \mathrm{J/mol},$$

and from $\Delta G = \Delta H - T\Delta S$:

$$\Delta S = \frac{\Delta H - \Delta G}{T} = \frac{1.5029\times10^5 - (-4.2145\times10^4)}{600} \approx 320.7\ \mathrm{J/K\,mol}.$$