Calculate - r G for the reaction at 27- CH 2 g -2 Ag - aq - 2 Ag s -2 H - aq Given - P H 2-0-5 bar- - Ag -10-5 M- H -10-3 M - - r G - Ag - aq -77-1 kJ mol -1- Given- ln2 - 104-9-90 A- -178-9 kJ mol -1B- -129-5 kJ mol -1C- -154-2 kJ mol -1D- None of these

The correct option is B 129.5 kJ mol^ 1Δ_r G^∘ = 0 77.1 × 2 = 154.2 kJ mol^ 1 Q = [H^+]^2P_H_2[Ag^+]^2 = 10^ 60.5 × (10^ 10) = 2 × 10^4 Δ G = Δ_r G^ ...

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Byju's Answer

Standard XII

Chemistry

Characteristics of Equilibrium Constant

Calculate Δ r...

Question

Calculate $Δ_{r} G$ for the reaction at $27^{\circ} C$
$H_{2} (g) + 2 A g^{+} (a q) ⇌ 2 A g (s) + 2 H^{+} (a q)$
Given : $P_{H_{2}} = 0.5 b a r; [A g^{+}] = 10^{- 5} M$
$[H^{+}] = 10^{- 3} M; Δ_{r} G^{\circ} [A g^{+} (a q)] = 77.1 k J m o l^{- 1}$
$(G i v e n, l n (2 \times 10^{4}) = 9.90)$

- 154.2 k J m o l^{- 1}

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- 178.9 k J m o l^{- 1}

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- 129.5 k J m o l^{- 1}

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None of these

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Solution

The correct option is C $- 129.5 k J m o l^{- 1}$
$Δ_{r} G^{\circ} = 0 - 77.1 \times 2 = - 154.2 k J m o l^{- 1} Q = \frac{[H^{+}]^{2}}{P_{H_{2}} [A g^{+}]^{2}} = \frac{10^{- 6}}{0.5 \times (10^{- 10})} = 2 \times 10^{4} Δ G = Δ_{r} G^{\circ} + R T l n Q Δ_{r} G = - 154.2 + 8.314 \times 10^{- 3} \times 300 l n (2 \times 10^{4}) = - 129.5 k J m o l^{- 1}$

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Similar questions

Q. Calculate

Δ_{r} G

for the reaction at

27^{\circ} C

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

Given :

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)

Q. Calculate

Δ_{r}

G for the reaction at

27^{\circ} C

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

Given :

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

Q. Calculate

Δ_{r} G

for the reaction at

27^{o} C

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

Given:

P_{H_{2}} = 0.5

bar;

[{A g}^{+}] = 10^{- 5} M

;

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

Q. Calculate

Δ_{r}

G for the reaction at

27^{\circ} C

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

Given :

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

Q. Calculate the equailibrium constant for the reaction:

{C u}_{(s)} + {2 A g}_{(a q)}^{+} ⟶ {C u}_{(a q)}^{2 +} + {2 A g}_{(s)}

Given that :

E_{A g^{+} / A g}^{0} = 0.80 V, E_{C u^{2 +} / C u}^{0} = 0.34 V .

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Calculate ΔrG for the reaction at 27∘C H2(g)+2Ag+(aq)⇌2Ag(s)+2H+(aq) Given : PH2=0.5 bar;[Ag+]=10−5M [H+]=10−3M;ΔrG∘[Ag+(aq)]=77.1 kJ mol−1 (Given, ln(2×104)=9.90)

The correct option is C −129.5 kJ mol−1ΔrG∘=0−77.1×2=−154.2 kJ mol−1Q=[H+]2PH2[Ag+]2=10−60.5×(10−10)=2×104ΔG=ΔrG∘+RT ln QΔrG=−154.2 + 8.314×10−3×300 ln (2×104)=−129.5 kJ mol−1

Calculate $Δ_{r} G$ for the reaction at $27^{\circ} C$
$H_{2} (g) + 2 A g^{+} (a q) ⇌ 2 A g (s) + 2 H^{+} (a q)$
Given : $P_{H_{2}} = 0.5 b a r; [A g^{+}] = 10^{- 5} M$
$[H^{+}] = 10^{- 3} M; Δ_{r} G^{\circ} [A g^{+} (a q)] = 77.1 k J m o l^{- 1}$
$(G i v e n, l n (2 \times 10^{4}) = 9.90)$