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Chemical Equilibrium question

2016 · Shift 0 · Q8
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Chemical Equilibrium question

2016 · Shift 0 · Q8

JEE MainChemistryChemical EquilibriumMCQ+4 / −1
The equilibrium constant at 298 K for a reaction A + B ⇋\leftrightharpoons⇋ C + D is 100. If the initial concentration of all the four species were 1M each, then equilibrium concentration of D (in mol L–1) will be:
  1. A
    0.818
  2. B
    1.818
  3. C
    1.182
  4. D
    0.182
View written solutionFree

Correct answer: B

  1. Write the equilibrium expression

For the reaction

A+B⇌C+DA + B \rightleftharpoons C + DA+B⇌C+D

we have

Kc=[C][D][A][B]=100K_c = \frac{[C][D]}{[A][B]} = 100Kc​=[A][B][C][D]​=100

at 298 K298\,\text{K}298K.

  1. Initial concentrations

Given initially:

[A]0=[B]0=[C]0=[D]0=1 M[A]_0 = [B]_0 = [C]_0 = [D]_0 = 1\,\text{M}[A]0​=[B]0​=[C]0​=[D]0​=1M

Since initially all are equal,

Qc=(1)(1)(1)(1)=1Q_c = \frac{(1)(1)}{(1)(1)} = 1Qc​=(1)(1)(1)(1)​=1

Because Qc<KcQ_c < K_cQc​<Kc​ (1<100)(1<100)(1<100), the reaction will proceed forward.

  1. Let the forward reaction proceed by xxx mol L−1^{-1}−1

Then at equilibrium:

[A]=1−x[A] = 1-x[A]=1−x [B]=1−x[B] = 1-x[B]=1−x [C]=1+x[C] = 1+x[C]=1+x [D]=1+x[D] = 1+x[D]=1+x

  1. Apply the equilibrium constant

Kc=(1+x)(1+x)(1−x)(1−x)=100K_c = \frac{(1+x)(1+x)}{(1-x)(1-x)} = 100Kc​=(1−x)(1−x)(1+x)(1+x)​=100

So,

(1+x)2(1−x)2=100\frac{(1+x)^2}{(1-x)^2} = 100(1−x)2(1+x)2​=100

Taking square root:

1+x1−x=10\frac{1+x}{1-x} = 101−x1+x​=10

  1. Solve for xxx

1+x=10(1−x)1+x = 10(1-x)1+x=10(1−x)

1+x=10−10x1+x = 10 - 10x1+x=10−10x

11x=911x = 911x=9

x=911=0.818…x = \frac{9}{11} = 0.818\ldotsx=119​=0.818…

  1. Find equilibrium concentration of DDD

[D]eq=1+x=1+911=2011=1.818 M[D]_{eq} = 1+x = 1 + \frac{9}{11} = \frac{20}{11} = 1.818\,\text{M}[D]eq​=1+x=1+119​=1120​=1.818M

  1. Check options
  • A: 0.8180.8180.818 → this is xxx, not [D][D][D]
  • B: 1.8181.8181.818 → correct
  • C: 1.1821.1821.182 → incorrect
  • D: 0.1820.1820.182 → incorrect

Therefore, the equilibrium concentration of DDD is

1.818 mol L−1\boxed{1.818\,\text{mol L}^{-1}}1.818mol L−1​

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