For a dimerization reaction, $2 \mathrm{~A}(\mathrm{~g}) \rightarrow \mathrm{A}_{2}(\mathrm{~g})$, at $298 \mathrm{~K}, \Delta \mathrm{U}^{\ominus}=-20 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta \mathrm{~S}^{\ominus}=-30 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, then the $\Delta \mathrm{G}^{\ominus}$ will be \_\_\_\_ J.

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Chemical Thermodynamics

For a dimerization reaction,

2 A (g) \to A_{2} (g)

,
at

298 K, Δ U^{⊖} = - 20 kJ mol^{- 1}, Δ S^{⊖} = - 30 JK^{- 1} mol^{- 1}

, then the

Δ G^{⊖}

will be ____ J.

Answer: -13537.57

First turn the internal-energy change into an enthalpy change. For

2 A (g) A X_{2} (g)

the gaseous moles drop from two to one, so

Δ n_{g} = - 1

and

Δ H = Δ U + Δ n_{g} R T = - 20000 + (- 1) (8.314) (298) = - 22477.57

J mol

^{- 1}

. Then

Δ G = Δ H - T Δ S = - 22477.57 - (298) (- 30) = - 22477.57 + 8940 = - 13537.57

J mol

^{- 1}

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Numerical answer