Chemistry - Chemical Equilibrium Question with Solution | TestHub

$\begin{array}{cc}& {\text{CH}}_3\text{COOH}\left(l\right)+{\text{C}}_2{\text{H}}_5\text{OH}\left(l\right)\rightleftharpoons {\text{C}\text{H}}_3{\text{COOC}}_2{\text{H}}_5\left(l\right)+{\text{H}}_2\text{O}\left(l\right)\\ \mathrm{Initial\; conc.}& \mathrm{ \; 1.00\; mol\; \; \; \; \; \; \; \; \; \; 0.180\; mol\; \; \; \; \; \; \; \; \; \; \; \; 0\; \; \; \; \; \; \; \; \; \; \; \; 0\; \; \; }\\ \mathrm{Equili.\; conc.}& \mathrm{ \; \; \; (1\; -\; x)\; mol\; \; \; \; \; \; \; (0.180\; -\; x)\; mol\; \; \; \; \; \; \; x\; mol\; \; \; \; \; \; \; \; \; x\; mol\; \; \; \; }\end{array}$

Given,$[{\mathrm{CH}}_3{\mathrm{COOC}}_2{\mathrm{H}}_5{]}_{\mathrm{equilibrium}} = 0.171 \mathrm{mol} = \mathrm{x}$

${\text{K}}_{\text{c}}=\frac{\left[{\text{CH}}_3{\text{COOC}}_2{\text{H}}_5\right]\left[{\text{H}}_2\text{O}\right]}{\left[{\text{CH}}_3\text{COOH}\right]\left[{\text{C}}_2{\text{H}}_5\text{OH}\right]}$

${\text{K}}_{\text{c}}=\frac{\text{0.171}\times \text{0.171}}{\left(1-\text{0.171}\right)\times \left(\text{0.180}-\text{0.171}\right)}$

${\text{K}}_{\text{c}}=\frac{\text{0.171}\times \text{0.171}}{\text{0.829}\times \text{0.009}}=\text{3.919}\approx \text{3.92}$