What is Kc for the following equilibrium when equilibrium concentration of each substance is [SO2] = 0.60 M, [O2] = 0.82 M and [SO3] = 1.90 M? 2SO2 (g)+O2(g) leftharpoons 2SO3 (g)

Question

What is Kc for the following equilibrium when equilibrium concentration of each substance is [SO2] = 0.60 M, [O2] = 0.82 M and [SO3] = 1.90 M? 2SO2 (g)+O2(g) leftharpoons 2SO3 (g)  (A) 12.229 (B) 21.229 (C) 12.922 (D) 21.046

Tardigrade Admin · Accepted Answer

For equation 2SO2 (g)+O2(g) leftharpoons 2SO3 (g) Kc=([SO3]2/[SO2]2 [O2]) =([1.90 M]2/[0.60 M]2 [0.82 M]) = 12.299 M-1 or 12.229 L mol-1

Q. What is $K_{c}$ for the following equilibrium when equilibrium concentration of each substance is $[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M$ and $[S O_{3}] = 1.90 M ?$
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$

12.229

21.229

12.922

21.046

For equation
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
$K_{c} = \frac{[ S O _{3} ] ^{2}}{[ S O _{2} ] ^{2} [ O _{2} ]}$
$= \frac{[ 1.90 M ] ^{2}}{[ 0.60 M ] ^{2} [ 0.82 M ]}$
$= 12.299 M^{- 1}$ or $12.229 L m o l^{- 1}$

Q. What is Kc​ for the following equilibrium when equilibrium concentration of each substance is [SO2​]=0.60M,[O2​]=0.82M and [SO3​]=1.90M? 2SO2​(g)+O2​(g)⇌2SO3​(g)