Equilibrium

Q1.

For a reaction; $X + Y ⇋ 2 Z$ , 1.0 mol of X, 1.5 mol of Y and 0.5 mol of Z were taken in a 1 L vessel and allowed to react. At equilibrium, the concentration of Z was 1.0 mol L^–1 . The equilibrium constant of the reaction is $\frac{x}{15}$ . The value of x is:

Explanation:

Hint: The formula of equilibrium constant is, $K_{equ} = \frac{Z^{2}}{X\times Y}$

Step 1:

The concentration of X, Y and Z is as follows:

Concentration = $=\frac{number \ of \ mole}{volume \ of \ solution}$

Hence, volume of solution is 1 L, concentration is equal to number of moles. So concentration of X, Y and Z is equal to its mole.

X + Y = 2Z
t = 0 1 1.5 0.5
At eq. 1-x 1.5-x 0.5 + 2x

The equilibrium concentration of Z is 1. Calculate the value of x as follows:

[Z] = 0.5 + 2x

$\frac{1-0.5}{2}$

x = 0.25 M

The concentration of X and Y at equilibrium is as follows:

[X] = 1 – x = 0.75, [Y] = 1.5 – x = 1.25

Step 2:

$K_{equ} = \frac{Z^{2}}{X\times Y}$

$K_{e q .} = \frac{1}{\frac{3}{4} \times \frac{5}{4}} = \frac{16}{15}$

Q2. The following equilibrium is established when $HCIO_4 $ is dissolved in weak acid HF.
$\mathrm{HF}+\mathrm{HClO}_4=\mathrm{ClO}_4^{-}+\mathrm{H}_2 \mathrm{~F}^{+} $
The correct set of conjugate acid-base pairs among the following is:

$\mathrm{HF}~\text{and}~ \mathrm{HClO}_4 $
$\mathrm{HF}~\text {and} ~\mathrm{ClO}_4^{-} $
$\mathrm{HF}\text { and }~\mathrm{H}_2 \mathrm{~F}^{+} $
$\mathrm{HClO}_4~ \text {and}~ \mathrm{H}_2 \mathrm{~F}^{+} $