Question

The equilibrium constant for the reaction
$C{H}_{3}COOH\left(aq\right)+C_2{H}_{5}OH\left(aq\right)\iff C{H}_{3}COO{C}_2{H}_5\left(aq\right)+H_{2}O\left(l\right)$ is 16. The initial concentration of each of $C{H}_{3}COOH$ and $C_2{H}_{5}OH$ is two (mole/lit). What is the concentration of $C{H}_{3}COO{C}_2{H}_5$ at equilibrium in moles / lit?

• A. 0.9
• B. 0.6
• C. 1.6
• D. 0.7

Answer 1

The correct option is C 1.6

The equilibrium expression is $C{H}_{3}COOH\left(aq\right)+C_2{H}_{5}OH\left(aq\right)\iff C{H}_{3}COO{C}_2{H}_5\left(aq\right)+H_{2}O\left(l\right)$
The expression for the equilibrium constant is $K_c=\frac{\left[C{H}_{3}COO{C}_2{H}_5\right]\left[H_{2}O\right]}{\left[C{H}_{3}COOH\right]\left[C_2{H}_{5}OH\right]}$
The following table lists the concentrations of various species.

	$C{H}_{3}COOH$	$C_2{H}_{5}OH$	$C{H}_{3}COO{C}_2{H}_5$	$H_{2}O$
Initial	2	2	0	0
Change	-x	-x	x	x
Equilibrium	2-x	2-x	x	x

Substitute values in the expression for the equilibrium constant
$16=\frac{x\times x}{(2-x)\times (2-x)}$
Hence, $x=1.6M$