At 1 atm pressure, ∆S=75 ${\rm{J}}{{\rm{K}}^{ - 1}}{\rm{mo}}{{\rm{l}}^{ - 1}}$ and ∆H=30 kJ ${\rm{mo}}{{\rm{l}}^{ - 1}}$.Find temperature of the reaction at equilibrium is:

NEET UG CHEMISTRY MCQ: THERMODYNAMICS

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Application +4 | -1

At 1 atm pressure, ∆S=75

\mathrm{J}{\mathrm{K}^{ - 1}}\mathrm{mo}{\mathrm{l}^{ - 1}}

and ∆H=30 kJ

\mathrm{mo}{\mathrm{l}^{ - 1}}

.Find temperature of the reaction at equilibrium is:

View answer and explanation

Correct option: A

400 K

Option A (400 K) is correct because the temperature at equilibrium is calculated using T = ∆H/∆S.

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{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