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

Question

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

Infinity Learn · Accepted Answer

Volume of vessel is doubled, concentrations become halved.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] $$=$$ $$10MAt$$ new equilibrium, [C] decreases whereas [A] & [B] increasesDue o doubling of volume [A] must be $$7.5$$ M but it is given that [A] $$=$$ $$10$$ MIncrease in concentration of 'A' is $$2.5$$ M and increase in concentration of 'B' must be $$2.5$$ MDecrease in concentration of 'C' must be $$2.5$$ MAt new equilibriumTemperature is constant, KC remains constant.Comparing (1) & (2)[C]eq in original equilibrium $$=$$ x $$=$$ $$45$$ M

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