MCQ
For a reaction at ${25\,^o}C$ enthalpy change and entropy changes are $ - 11.7 \times {10^3}\,J\,mo{l^{ - 1}}$ and $ - 105\,J\,mo{l^{ - 1}}\,{K^{ - 1}}$ respectively. What is the Gibbs free energy .....$kJ$
- A$15.05$
- ✓$19.59$
- C$2.55$
- D$22.55$
✓
Answer
Correct option: B.
$19.59$
(b)$\Delta G = \Delta H - T\Delta S,\,\,\,T = 25 + 273 = 298\,K$
$ = - 11.7 \times {10^3} - 298 \times ( - 105) = 19590\,J = 19.59\,kJ$
$ = - 11.7 \times {10^3} - 298 \times ( - 105) = 19590\,J = 19.59\,kJ$
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