For CH3COOH + C2H5OH CH3COOC2H5 + H2O in solution, K_c = 4.0. A mixture has [CH3COOH] = 0.20, [CH3COOC2H5] = 0.30 and [H2O] = 0.45 (all in mol L^-1). The net reaction goes forward only if [C2H5OH] exceeds a critical value. Find this value in mol L^-1 (two decimals).

JEE Main levelNumerical answerDifficulty 2 of 3ce127

For $CH X_{3} COOH + C X_{2} H X_{5} OH CH X_{3} COOC X_{2} H X_{5} + H X_{2} O$ in solution, $K_{c} = 4.0$ . A mixture has $[CH X_{3} COOH] = 0.20$ , $[CH X_{3} COOC X_{2} H X_{5}] = 0.30$ and $[H X_{2} O] = 0.45$ (all in $mol L^{- 1}$ ). The net reaction goes forward only if $[C X_{2} H X_{5} OH]$ exceeds a critical value. Find this value in $mol L^{- 1}$ (two decimals).

Numerical answer type, answer in mol/L, tolerance 0.01.

Forward reaction needs

Q_{c} < K_{c}

\frac{( 0.30 ) ( 0.45 )}{( 0.20 ) [ C X _{2} H X _{5} OH ]} < 4.0

. Hence

[C X_{2} H X_{5} OH] > \frac{0.135}{0.20 \times 4.0} = 0.1688

, i.e.

0.17 mol L^{- 1}

to two decimals.