Question

When one mole of benzoic acid $\left(C_6{H}_{5}COOH\right)$ and three moles of ethanol $\left(C_2{H}_{5}OH\right)$ are mixed and kept at ${200}^{0}C$ until equilibrium is reached, it is found that 87% of the acid is consumed by the reaction.
$C_6{H}_{5}COOH\left(l\right)+C_2{H}_{5}OH\left(l\right)\rightleftharpoons C_6{H}_{5}COO{C}_2{H}_5\left(l\right)+H_{2}O\left(l\right)$
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:

Answer 1

$C_6{H}_{5}COOH\left(l\right)+C_2{H}_{5}OH\left(l\right)\rightleftharpoons C_6{H}_{5}COO{C}_2{H}_5\left(l\right)+H_{2}O\left(l\right)$ at ${200}^{0}C$

At $1$ $3$ $0$ $-$

$t=0$

At $1-\alpha$ $3(1-\alpha )$ $\alpha$ $-$

Given $87$% of acid consumed

So $\alpha =0.87$

$K_C=\frac{0.87}{3\times 0.13\times 0.13}=17.16\to \left(1\right)$

Now $4$ moles of ethanol & $1$ mole benzoic acid

$C_6{H}_{5}COOH+C_2{H}_{5}OH\rightleftharpoons C_6{H}_{5}COO{C}_2{H}_5+H_{2}O$

At $1$ $4$ $0$ $-$

$t=0$

At $1-{\alpha }^1$ $4(1-{\alpha }^1)$ ${\alpha }^1$

$t=t_{eq}$

$K_C=\frac{{\alpha }^1}{4{(1-{\alpha }^1)}^2}\to \left(2\right)$

$K_C$ doesn't depend on initial concentration

So equation (1) & equation (2) are equal

$\frac{{\alpha }^1}{{(1-{\alpha }^1)}^2\times 4}=17.16$

${\alpha }^1=({\alpha }^1-{2\alpha }^1+1)\left(68.64\right)$

${\alpha }^{1^2}\left(68.64\right)-\left(138.28\right){\alpha }^1+68.64=0$

${\alpha }^1=0.8863$ Hence $88.63$% acid dissociated