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Question

# Calculate the equilibrium concentration ratio of C to A, if 2 moles each of A and B are allowed to come to equilibrium at 300 K. A (g)+B (g)⇌C (g)+D (g) ΔG∘=−1380 cal/mol

A
0.56
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B
4.45
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C
1.50
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D
3.16
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Solution

## The correct option is D 3.16We know , ΔG=ΔG∘+2.303 RT logQ where: ΔG,ΔG∘ are Gibbs free energy and standard gibbs free energy respectively. R is the universal gas constant T is the temperature Q is the reaction quotient at time t At equilibrium, ΔG=0, Q=Kc∴ ΔG∘=−2.303 RT logKc−1380=−2.303×2×300×logKclogKc=−1380−2.303×2×300=23002303⇒logKc=0.99≈1∴logKc=1Kc=101=10 Given reaction : A (g)+B (g)⇌C (g)+D (g)Initial(mol) 2 2 0 0Equilibrium: 2−α 2−α α αKc=[C]eq[D]eq[A]eq[B]eq Taking the volume V Equilibrium ratio of C to A is : [C]eq[A]eq=[αV][(2−α)V]=α2−α Kc=[αV]2[(2−α)V]2=α2(2−α)210=α2(2−α)2√10=α2−α=[C]eq[A]eq[C]eq[A]eq=√10=3.16

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