The equilibrium composition for the reaction If 0.2 mol of is added at the same temperature, the equilibrium concentration of is Given: for the reaction at 298 K is 20. [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
The equilibrium composition for the reaction

${PCl}_{3} + {Cl}_{2} ⇌ {PCl}_{5} at 298 K is given below:$

${[{PCl}_{3}]}_{eq} = {0.2 mol L}^{- 1}, {[{Cl}_{2}]}_{eq} = {0.1 mol L}^{- 1}, {[{PCl}_{5}]}_{eq} = {0.40 mol L}^{- 1}$

If 0.2 mol of ${Cl}_{2}$ is added at the same temperature, the equilibrium concentration of ${PCl}_{5}$ is $_____\times 10^{- 2} {mol L}^{- 1}$

Given: $K_{c}$ for the reaction at 298 K is 20. [2023]

Ans.
(49)

$\begin{matrix} {PCl}_{3} (g) & + & {Cl}_{2} (g) & ⇌ & {PCl}_{5} (g) \\ At equilibrium & 0.2 \frac{mol}{Lit} & 0.1 \frac{mol}{Lit} & 0.4 \frac{mol}{Lit} \\ New equilibrium & (0.1 - x) & (0.3 - x) & (0.4 + x) \end{matrix}$

$K_{c} = (\frac{0.4 + x}{(0.2 - x) (0.3 - x)}) \Rightarrow 20 = \frac{0.4 + x}{(0.2 - x) (0.3 - x)}$

$(0.4 + x) = 20 (0.3 - x) (0.2 - x)$

$∴$    $0.4 + x = 20 (0.06 + x^{2} - 0.5 x)$

   $20 x^{2} - 11 x + 0.8 = 0$

   $x = \frac{11 \pm \sqrt{121 - 64}}{40}$

   $x \approx 0.08625$

$∴$ $({PCl}_{5}) = 0.48625 \approx 49 \times 10^{- 2}$

Want Us to Email you About Special Offers & Updates?