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Chemical Equilibrium question

2021 · 25 Jul · Shift 1 · Q19
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  5. /2021 · 25 Jul · Shift 1 · Q19

Chemical Equilibrium question

2021 · 25 Jul · Shift 1 · Q19

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
For the reaction A + B ⇌\rightleftharpoons⇌ 2C the value of equilibrium constant is 100 at 298 K. If the initial concentration of all the three species is 1 M each, then the equilibrium concentration of C is x ×\times× 10 −-− 1 M. The value of x is ‾\underline{\hspace{2cm}}​. (Nearest integer)
Numerical answer
View written solutionFree

Correct answer: 25

  1. Write the equilibrium expression

For A+B⇌2CA + B \rightleftharpoons 2CA+B⇌2C

the equilibrium constant is Kc=[C]2[A][B]=100K_c = \frac{[C]^2}{[A][B]} = 100Kc​=[A][B][C]2​=100

  1. Initial concentrations

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

Let the reaction proceed forward by amount yyy. Then at equilibrium: [A]=1−y[A] = 1-y[A]=1−y [B]=1−y[B] = 1-y[B]=1−y [C]=1+2y[C] = 1+2y[C]=1+2y

  1. Apply the equilibrium constant

Substitute into the expression: (1+2y)2(1−y)(1−y)=100\frac{(1+2y)^2}{(1-y)(1-y)} = 100(1−y)(1−y)(1+2y)2​=100

So, (1+2y)2(1−y)2=100\frac{(1+2y)^2}{(1-y)^2} = 100(1−y)2(1+2y)2​=100

Taking square root: 1+2y1−y=10\frac{1+2y}{1-y} = 101−y1+2y​=10

Therefore, 1+2y=10−10y1+2y = 10-10y1+2y=10−10y 12y=912y = 912y=9 y=34y = \frac{3}{4}y=43​

  1. Find equilibrium concentration of CCC

[C]eq=1+2y=1+2(34)=1+32=52=2.5 M[C]_{eq} = 1+2y = 1+2\left(\frac{3}{4}\right) = 1+\frac{3}{2} = \frac{5}{2} = 2.5\,\text{M}[C]eq​=1+2y=1+2(43​)=1+23​=25​=2.5M

Given that equilibrium concentration of CCC is written as x×10−1 Mx \times 10^{-1}\,\text{M}x×10−1M

So, x×10−1=2.5x \times 10^{-1} = 2.5x×10−1=2.5 x=25x = 25x=25

  1. Compare with stored answer

Derived answer = 252525

Stored correct answer = 252525

They match.

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