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For $A + B ⇌ C$ , the equilibrium concentrations of A and B at a temperature are 15 mole/lit. When volume is doubled the reaction has equilibrium concentration of A as 10 mol/lit. The concentration of C in original equilibrium is

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45 M

15 M

10 M

30 M

answer is D.

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Detailed Solution

\large \mathop {}\limits_{At{\text{ }}equilibrium} \,\mathop A\limits_{15M} + \mathop B\limits_{15M} \rightleftharpoons \mathop C\limits_x

\large {K_C} = \frac{{\left[ C \right]}}{{\left[ A \right]\left[ B \right]}} = \frac{x}{{225}}{M^{ - 1}} \to \left( 1 \right)

Volume of vessel is doubled, concentrations become halved.

\large \therefore \left[ A \right] = \frac{{15}}{2}M = \left[ B \right];\left[ C \right] = \frac{x}{2}

But according to Lechatelier's principle increase in volume favours reaction involving increase in numberr of moles. Thus reaction proceeds in backward direction at new equilibrium is attained.

At new equilibrium, [A] = 10M

At new equilibrium, [C] decreases whereas [A] & [B] increases

Due o doubling of volume [A] must be 7.5 M but it is given that [A] = 10 M

Increase in concentration of 'A' is 2.5 M and increase in concentration of 'B' must be 2.5 M

Decrease in concentration of 'C' must be 2.5 M

At new equilibrium

\large \therefore \mathop {A(g)}\limits_{\left( {7.5 + 2.5} \right)M} + \mathop {B(g)}\limits_{\left( {7.5 + 2.5} \right)M} \rightleftharpoons \mathop {K(g)}\limits_{\left( {\frac{x}{2} - 2.5} \right)}

\large {K_C} = \frac{{\left( {\frac{{x - 5}}{2}} \right)}}{{\left( {10} \right)\left( {10} \right)}} \to \left( 2 \right)

Temperature is constant, K_C remains constant.

Comparing (1) & (2)

\large \frac{x}{{225}} = \frac{{x - 5}}{{200}}

[C]_eq in original equilibrium = x = 45 M

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