Question:

The variation of equilibrium constant with temperature is given below :
TemperatureEquilibrium Constant
T1 = 25\(^\circ\)CK1 = 10
T2 = 100\(^\circ\)CK2 = 100
The values of \(\Delta H^\circ\), \(\Delta G^\circ\) at T1 and \(\Delta G^\circ\) at T2 (in kJ mol-1) respectively, are close to :
[Use R = 8.314 J K-1 mol-1]

Show Hint

Use the van't Hoff equation \(\ln\frac{K_2}{K_1}=\frac{\Delta H^\circ}{R}\left(\frac1{T_1}-\frac1{T_2}\right)\) for \(\Delta H^\circ\) and \(\Delta G^\circ=-RT\ln K\) at each temperature.
Updated On: Nov 5, 2026
  • 28.4, \(-5.71\) and \(-14.29\)
  • 0.64, \(-7.14\) and \(-5.71\)
  • 28.4, \(-7.14\) and \(-5.71\)
  • 0.64, \(-5.71\) and \(-14.29\)
Show Solution
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The Correct Option is A

Solution and Explanation

T1 = 298 K, T2 = 373 K.
van't Hoff: \(\ln\frac{K_2}{K_1}=\ln10=2.303=\frac{\Delta H^\circ}{8.314}\left(\frac{1}{298}-\frac{1}{373}\right)\).
\(\frac{1}{298}-\frac{1}{373}=\frac{75}{298\times373}=6.75\times10^{-4}\ \mathrm{K^{-1}}\).
\(\Delta H^\circ=\frac{8.314\times2.303}{6.75\times10^{-4}}\approx2.84\times10^{4}\) J/mol = 28.4 kJ/mol.
\(\Delta G^\circ_{T_1}=-RT_1\ln K_1=-8.314\times298\times2.303=-5705\) J = \(-5.71\) kJ/mol.
\(\Delta G^\circ_{T_2}=-RT_2\ln K_2=-8.314\times373\times4.606\approx-14.28\times10^{3}\) J, i.e. about \(-14.29\) kJ/mol.
Answer: (1) 28.4, \(-5.71\) and \(-14.29\)
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