Compute -rG for the reaction H2O - 1 atm- 323 K - H2O g- 1 atm- 323 KGiven that - -vapH at 373 K - 40-639 kJmol1- CPH2O- l - 75-312 J K1 mol1-CPH2O- g - 33-305 J K1mol1

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Gibbs Free Energy & Spontaneity

Compute ΔrG...

Question

Compute $Δ_{r} G$ for the reaction $H_{2} O (;, 1 a t m, 323 K) \to H_{2} O (g, 1 a t m, 323 K)$
Given that : $Δ_{v a p} H$ at $373 K = 40.639 k J m o l^{1}, C_{P} (H_{2} O, l) = 75.312 J K^{1} m o l^{1}, C_{P} (H_{2} O, g) = 33.305 J K^{1} m o l^{1}$

Δ_{r} G = 5.59 k J m o l^{1}

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Δ_{r} G = - 5.59 k J m o l^{1}

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Δ_{r} G = 55.9 k J m o l^{1}

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None of these

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Δ_{r} G

27^{\circ} C

H_{2} (g) + 2 A g^{+} (a q) ⇌ 2 A g (s) + 2 H^{+} (a q)

P_{H_{2}} = 0.5 b a r; [A g^{+}] = 10^{- 5} M

[H^{+}] = 10^{- 3} M; Δ_{r} G^{\circ} [A g^{+} (a q)] = 77.1 k J m o l^{- 1}

(G i v e n, l n (2 \times 10^{4}) = 9.90)

Which quantity out of $Δ_{r} G$ and $Δ_{r} G^{⊖}$ will be zero at equilibrium?

Δ_{r} G

27^{\circ} C

H_{2} (g) + 2 A g^{+} (a q) ⇌ 2 A g (s) + 2 H^{+} (a q)

P_{H_{2}} = 0.5 b a r; [A g^{+}] = 10^{- 5} M

[H^{+}] = 10^{- 3} M; Δ_{r} G^{\circ} [A g^{+} (a q)] = 77.1 k J m o l^{- 1}

(G i v e n, l n (2 \times 10^{4}) = 9.90)

Δ_{r}

27^{\circ} C

H_{2} (g) + 2 A g^{+} (a q) ⇌ 2 A g (s) + 2 H^{+} (a q)

p_{H_{2}}

0.5 bar; [A g^{+}] = 10^{- 5} M; log 2 =0.3 log 2= 0.7

[H^{+}] = 10^{- 3} M; Δ_{r} G^{\circ} [A g^{+} (a q)] = 77.1 kJ/mol

A_{2} g + B_{2} (g) ⇌ 2 A B (g)

Δ_{r} G^{\circ} a n d Δ_{r} S^{\circ}

J K^{- 1} m o l^{- 1}

Δ_{r} C_{P} i s 20 J K^{- 1} m o l^{- 1}

\frac{Δ_{r} H^{\circ}}{10}

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