For a reaction at 25℃, enthalpy and entropy changes are $ - 11.7{\rm{ \times }}{10^3}Jmo{l^{ - 1}}$ and $ - 105Jmo{l^{ - 1}}{K^{ - 1}}$respectively. What is the Gibbs free energy?

NEET UG CHEMISTRY MCQ: THERMODYNAMICS

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Application +4 | -1

For a reaction at 25℃, enthalpy and entropy changes are

- 11.7\mathrm{ \times }{10^3}Jmo{l^{ - 1}}

and

- 105Jmo{l^{ - 1}}{K^{ - 1}}

respectively. What is the Gibbs free energy?

View answer and explanation

Correct option: B

19.59 kJ

Option B (19.59 kJ) is correct because the Gibbs free energy is calculated using ∆G = ∆H - T∆S.

Practice All Chapter MCQs →

{T_i}

{T_f}

\mathrm{H}_2 \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})

\mathrm{X}_2 \mathrm{O_4(l}) \longrightarrow 2 \mathrm{XO}_2(\mathrm{g}), \Delta U=2.1 \mathrm{~kcal}, \Delta S=20 \mathrm{~cal} \mathrm{~K}^{-1}

300 \mathrm{~K}

\Delta G