Calculate - r G for the reaction at 27- C H 2 g -2 Ag - aq - 2 Ag s -2 H - aq Given - PH 2-0-5 bar- - Ag -10-5 M - log 2-0-3 log 2-0-7- H -10-3 M - - r G - Ag - aq -77-1 kJ - molA- -154-20 kJ - molB- -178-90 kJ - molC- -129-50 kJ - molD- -154-21 kJ - mol

The correct option is C 129.50 kJ/molΔ_r G^o = 0 77.1×2 = 154.2 kJmol^ 1 Q= [H^+]^2/P_H_2[Ag^+]^2 Q= 10^ 6/0.5×10^ 10 Q= 2×10^4 Δ_r G = nbsp;Δ_r G^o nbs ...

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Byju's Answer

Standard XII

Chemistry

Equilibrium Constant and Standard Free Energy Change

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 : $p_{H_{2}}$ = $0.5 bar; [A g^{+}] = 10^{- 5} M; log 2 =0.3 log 2= 0.7$ ,
$[H^{+}] = 10^{- 3} M; Δ_{r} G^{\circ} [A g^{+} (a q)] = 77.1 kJ/mol$

- 154.20 kJ/mol

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- 178.90 kJ/mol

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- 129.50 kJ/mol

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+ 154.21 kJ/mol

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Solution

The correct option is C $- 129.50 kJ/mol$
$Δ_{r} G^{o} = 0 - 77.1 \times 2 = - 154.2 k J m o l^{- 1}$
$Q = \frac{[H^{+}]^{2}}{P_{H_{2}} [A g^{+}]^{2}}$
$Q = \frac{10^{- 6}}{0.5 \times 10^{- 10}}$
$Q = 2 \times 10^{4}$
$Δ_{r} G = Δ_{r} G^{o} + R T l n Q$
$Δ_{r} G = - 154.2 + \frac{8.314 \times 300 \times l n (2 \times 10^{4})}{1000}$
$Δ_{r} G = - 129.5 k J m o l^{- 1}$

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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 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 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 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. Consider the following cell reaction:

2 A g C l (s) + H_{2} (g) \to 2 A g (s) + 2 H^{+} (a q) + 2 C l^{-} (a q); E^{o} = 0.19 V

[C l^{-}] = 10^{- 3} M

and

P_{H_{2}} = 0.01 atm

, if the cell potential is 0.43

V

25^{o} C

the

p H

is:

(Take \frac{2.303 R \times 298}{F} = 0.06)

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Calculate Δr G for the reaction at 27∘C H2(g)+2Ag+(aq)⇌2Ag(s)+2H+(aq) Given : pH2 = 0.5 bar;[Ag+]=10−5M;log 2 =0.3 log 2= 0.7, [H+]=10−3M;ΔrG∘[Ag+(aq)]=77.1 kJ/mol

The correct option is C −129.50 kJ/molΔr Go=0−77.1×2=−154.2 kJmol−1 Q= [H+]2PH2[Ag+]2 Q= 10−60.5×10−10 Q= 2×104 Δr G=Δr Go+ RTlnQ Δr G=−154.2+8.314×300×ln(2× 104)1000 ​Δr G=−129.5kJmol−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 : $p_{H_{2}}$ = $0.5 bar; [A g^{+}] = 10^{- 5} M; log 2 =0.3 log 2= 0.7$ ,
$[H^{+}] = 10^{- 3} M; Δ_{r} G^{\circ} [A g^{+} (a q)] = 77.1 kJ/mol$