What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2]= 0.60M, [O2] = 0.82M and [SO3] = 1.90M ?

Asked by Abhisek | 1 year ago |  82

Solution :-

As per the question,

$$2SO_{2}(g)+O_{2}(g) ⇌ 2SO_{3}(g)(Given)$$

$$K_{c}=\dfrac{[SO_{3}]^{2}}{[SO_{2}]^{2}[O_{2}]}\\ \\$$

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

$$=12.229M^{-1}(approximately)$$

Hence, K for the equilibrium is 12.229 M–1.

Answered by Pragya Singh | 1 year ago

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