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What is \mathrm{Kc} for the following equilibrium when the equilibrium concentration of each substance is: \left[\mathrm{SO}_{2}\right]=0.60 \mathrm{M},\left[\mathrm{O}_{2}\right]=0.82 \mathrm{M} and \left[\mathrm{SO}_{3}\right]=1.90 \mathrm{M} ? 2 S O_{2}(g)+O_{2}(g) \rightleftharpoons 2 S O_{3}(g)
CBSE | Chemistry | Class 11 | Equilibrium | NCERT
What is \mathrm{Kc} for the following equilibrium when the equilibrium concentration of each substance is: \left[\mathrm{SO}_{2}\right]=0.60 \mathrm{M},\left[\mathrm{O}_{2}\right]=0.82 \mathrm{M} and \left[\mathrm{SO}_{3}\right]=1.90 \mathrm{M} ? 2 S O_{2}(g)+O_{2}(g) \rightleftharpoons 2 S O_{3}(g)

Answer:

Given information in the question,

2 S O_{2}(g)+O_{2}(g) \rightleftharpoons 2 S O_{3}(g)
As we know that,

K_{c}=\frac{\left[S O_{3}\right]^{2}}{\left[S O_{2}\right]^{2}\left[O_{2}\right]}
Thus, evaluating by substiuting the value of given parameters,

=\frac{(1.9)^{2} M^{2}}{(0.6)^{2}(0.82) M^{3}}

=12.229 M^{-1} (approximately)

Hence, K for the equilibrium is 12.229 \mathrm{M}^{-1},

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