What is Kc the following equilibrium when the equilibrium concentration of each substance is -SO-2-0-60M- -O-2-0-82M- -SO-3-1-90M quad 2SO-2-O-2- rightleftharpoons 2SO-3-

Question

Toppr Admin · Accepted Answer

2SO2-O2-x21CC-2SO3KC-SO3-2-SO2-2-O2-1-90-2-0-60-2-0-82-12-23

What is $K c$ for the following equilibrium when the equilibrium concentration of each substance is $[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M, [S O_{3}] = 1.90 M 2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ .

${2 S O}_{2} + O_{2} ⇌ {2 S O}_{3}$
$K_{C} = \frac{{[{S O}_{3}]}^{2}}{{[{S O}_{2}]}^{2} [O_{2}]} = \frac{{(1.90)}^{2}}{{(0.60)}^{2} (0.82)} = 12.23$

What is Kc for the following equilibrium when the equilibrium concentration of each substance is [SO2]=0.60M,[O2]=0.82M,[SO3]=1.90M2SO2+O2⇌2SO3.

2SO2+O2⇌2SO3KC=[SO3]2[SO2]2[O2]=(1.90)2(0.60)2(0.82)=12.23

What is $K c$ for the following equilibrium when the equilibrium concentration of each substance is $[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M, [S O_{3}] = 1.90 M 2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ .

${2 S O}_{2} + O_{2} ⇌ {2 S O}_{3}$
$K_{C} = \frac{{[{S O}_{3}]}^{2}}{{[{S O}_{2}]}^{2} [O_{2}]} = \frac{{(1.90)}^{2}}{{(0.60)}^{2} (0.82)} = 12.23$