JEE PYQ: Thermodynamics Question 11
Question 11 - 2020 (06 Sep Shift 2)
The variation of equilibrium constant with temperature is given below:
| Temperature | Equilibrium Constant |
|---|---|
| $T_1 = 25$ $^\circ$C | $K_1 = 10$ |
| $T_2 = 100$ $^\circ$C | $K_2 = 100$ |
The values of $\Delta H^\circ$, $\Delta G^\circ$ at $T_1$ and $\Delta G^\circ$ at $T_2$ (in kJ mol$^{-1}$), respectively, are close to
[use $R = 8.314$ J K$^{-1}$ mol$^{-1}$]
(1) 28.4, $-7.14$ and $-5.71$
(2) 0.64, $-7.14$ and $-5.71$
(3) 28.4, $-5.71$ and $-14.29$
(4) 0.64, $-5.71$ and $-14.29$
Show Answer
Answer: (3)
Solution
$\Delta G^\circ = -RT\ln K$. $\Delta G^\circ$ at $T_1 = -8.314 \times 298 \times 2.303 \times \log 10 = -5.71$ kJ/mol. $\Delta G^\circ$ at $T_2 = -8.314 \times 298 \times 373 \times 2.303 \times \log(100) = -14.29$ kJ/mol. $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. $-5.71 = \Delta H^\circ - 298(\Delta S^\circ)$. $-14.29 = \Delta H^\circ - 373(\Delta S^\circ)$. $\Delta H^\circ = 28.4$ kJ/mol.
BETA
