For the above reaction at 298 K, is found to be . If the concentration of at equilibrium is 0.040 M, then the concentration of in M is [2022]

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.
$3 O_{2 (g)} ⇌ 2 O_{3 (g)}$

For the above reaction at 298 K, $K_{c}$ is found to be $3.0 \times 10^{- 59}$ . If the concentration of $O_{2}$ at equilibrium is 0.040 M, then the concentration of $O_{3}$ in M is [2022]

1 $4.38 \times 10^{- 32}$

2 $1.9 \times 10^{- 63}$

3 $2.4 \times 10^{31}$

4 $1.2 \times 10^{21}$

Ans.
(1)

$\begin{matrix} 3 O_{2 (g)} & ⇌ & 2 O_{3 (g)} \\ Initial conc . & 1 & 0 \\ At equilibrium & 1 - 3 x & 2 x \end{matrix}$

$K_{c} = \frac{{[O_{3}]}^{2}}{{[O_{2}]}^{3}} = 3 \times 10^{- 59}$

Given: $[O_{2}] = 0.040$ M

$K_{c} = \frac{{[O_{3}]}^{2}}{{(0.040)}^{3}} = 3 \times 10^{- 59}$

${[O_{3}]}^{2} = 1.92 \times 10^{- 63}$

$[O_{3}] = 4.38 \times 10^{- 32}$ M

Want Us to Email you About Special Offers & Updates?