What is Kc for the following equilibrium when equilibrium concentration of each substance is [8 [SO2]=0.60M, [O2]=0.82 M and [SO3]=1.90M ? 2SO2(g)+O2(g)leftharpoons 2SO3(g)

Question

What is Kc for the following equilibrium when equilibrium concentration of each substance is [8 [SO2]=0.60M, [O2]=0.82 M and [SO3]=1.90M ? 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/[0.60M]2[0.82M]) =12.229 M-1 or 12.229 L mol-1

Q. What is $K_{c}$ for the following equilibrium when equilibrium concentration of each substance is [8 $[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M$ 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 $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ $K_{c} = \frac{[ S O _{3} ] ^{2}}{[ S O _{2} ] ^{2} [ O _{2} ]}$ $= \frac{[ 1.90 M ] ^{2}}{[ 0.60 M ] ^{2} [ 0.82 M ]}$ $= 12.229 M^{- 1}$ or $12.229 L m o l^{- 1}$

Q. What is Kc​ for the following equilibrium when equilibrium concentration of each substance is [8 [SO2​]=0.60M,[O2​]=0.82M and [SO3​]=1.90M ? 2SO2​(g)+O2​(g)⇌2SO3​(g)