Question

At \(298\,K\), for the reaction \[ N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g) \] \[ \Delta H=-92.4\ \text{kJ} \] and \[ \log K_c=5.75. \] What is the entropy change \((\Delta S)\) (in J K\(^{-1}\)) for this reaction at the same temperature? \[ (R=8.3\ \text{J K}^{-1}\text{mol}^{-1}) \]

• A. \(-300\)
• B. \(-400\)
• C. \(-200\)
• D. \(-100\)

Hint:

Useful relations: \[ \Delta G=-RT\ln K \] \[ \Delta G=\Delta H-T\Delta S \] and \[ \ln K=2.303\log K. \] For the Haber process, entropy decreases because \[ 4\ \text{moles of gas} \rightarrow 2\ \text{moles of gas}. \]

Answer 1

The Correct Option is C

Concept: \[ \Delta G=\Delta H-T\Delta S \] and \[ \Delta G=-RT\ln K. \] Hence, \[ \Delta H-T\Delta S=-RT\ln K. \] \[ \Delta S=\frac{\Delta H+RT\ln K}{T}. \]

Step 1: Calculate \(\ln K\). Given, \[ \log K_c=5.75. \] Using \[ \ln K=2.303\log K, \] \[ \ln K = 2.303\times5.75 = 13.242. \]

Step 2: Calculate \(\Delta G\). \[ \Delta G = -RT\ln K. \] \[ = -(8.3)(298)(13.242). \] \[ = -32771\ \text{J} \] \[ = -32.77\ \text{kJ}. \]

Step 3: Use \(\Delta G=\Delta H-T\Delta S\). \[ -32.77 = -92.4-T\Delta S. \] \[ T\Delta S = -92.4+32.77. \] \[ T\Delta S = -59.63\ \text{kJ}. \] \[ \Delta S = \frac{-59.63\times10^3}{298}. \] \[ \Delta S = -200.1\ \text{J K}^{-1}. \]

Final Answer: \[ \boxed{\Delta S\approx -200\ \text{J K}^{-1}} \] \[ \boxed{\text{Answer = (C)}} \]