3 O 2( g ) leftharpoons 2 O 3( g ) for the above reaction at 298 K , K c is found to be 3.0 × 10-59. If the concentration of O 2 at equilibrium is 0.040 M then concentration of O 3 in M is

Question

3 O 2( g ) leftharpoons 2 O 3( g ) for the above reaction at 298 K , K c is found to be 3.0 × 10-59. If the concentration of O 2 at equilibrium is 0.040 M then concentration of O 3 in M is (A) 1.2 × 1021 (B) 4.38 × 10-32 (C) 1.9 × 10-63 (D) 2.4 × 1031

Tardigrade Admin · Accepted Answer

3 O 2( g ) leftharpoons 2 O 3( g ) K c =([ O 3]2/[ O 2]3) 3 × 10-59=([ O 3]2/(4 × 10-2)3) [ O 3]2=3 × 10-59 × 64 × 10-6 =19.2 × 10-64 =4.38 × 10-32 M

Q. $3 O_{2} (g) ⇌ 2 O_{3} (g)$
for the above reaction at $298 K, K_{c}$ is found to be $3.0 \times 1 0^{- 59}$ . If the concentration of $O_{2}$ at equilibrium is $0.040 M$ then concentration of $O_{3}$ in $M$ is

$1.2 \times 1 0^{21}$

$4.38 \times 1 0^{- 32}$

$1.9 \times 1 0^{- 63}$

$2.4 \times 1 0^{31}$

$3 O_{2} (g) ⇌ 2 O_{3} (g)$
$K_{c} = \frac{[ O _{3} ] ^{2}}{[ O _{2} ] ^{3}}$
$3 \times 1 0^{- 59} = \frac{[ O _{3} ] ^{2}}{( 4 \times 1 0 ^{- 2} ) ^{3}}$
$[O_{3}]^{2} = 3 \times 1 0^{- 59} \times 64 \times 1 0^{- 6}$
$= 19.2 \times 1 0^{- 64}$
$= 4.38 \times 1 0^{- 32} M$

Q. 3O2​(g)⇌2O3​(g) for the above reaction at 298K,Kc​ is found to be 3.0×10−59. If the concentration of O2​ at equilibrium is 0.040M then concentration of O3​ in M is

1.2×1021

4.38×10−32

1.9×10−63

2.4×1031