When one mole of benzoic acid $(C_{6} H_{5} C O O H)$ and three moles of ethanol $(C_{2} H_{5} O H)$ are mixed and kept at $200^{o} C$ until equilibrium is reached, it is found that 87% of the acid is consumed by the reaction
$C_{6} H_{5} C O O H (l) + C_{2} H_{5} O H (l) ⇌ C_{6} H_{5} C O O C_{2} H_{5} (l) + H_{2} O$
Find out the percentage of the acid consumed when one mole of the benzoic acid is mixed with four moles of ethanol and treated in the same way.

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Solution

$C_{6} H_{5} C O O H (l) + C_{2} H_{5} O H (l) ⇌ C_{6} H_{5} C O O C_{2} H_{5} (l) + H_{2} O (l)$ at $200^{0} C$
At $1$ $3$ $0$ $-$
$t = 0$
At $1 - α$ $3 (1 - α)$ $α$ $-$
Given $87$ % of acid consumed
So $α = 0.87$
$K_{C} = \frac{0.87}{3 \times 0.13 \times 0.13} = 17.16 \to (1)$
Now $4$ moles of ethanol & $1$ mole benzoic acid
$C_{6} H_{5} C O O H + C_{2} H_{5} O H ⇌ C_{6} H_{5} C O O C_{2} H_{5} + H_{2} O$
At $1$ $4$ $0$ $-$
$t = 0$
At $1 - α^{1}$ $4 (1 - α^{1})$ $α^{1}$
$t = t_{e q}$
$K_{C} = \frac{α^{1}}{4 {(1 - α^{1})}^{2}} \to (2)$
$K_{C}$ doesn't depend on initial concentration
So equation (1) & equation (2) are equal
$\frac{α^{1}}{{(1 - α^{1})}^{2} \times 4} = 17.16$
$α^{1} = (α^{1} - {2 α}^{1} + 1) (68.64)$
$α^{1^{2}} (68.64) - (138.28) α^{1} + 68.64 = 0$
$α^{1} = 0.8863$ Hence $88.63$ % acid dissociated

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