What is `K_(c)` for the following equilibrium when the equilibrium concentration of each substance is: `[SO_(2)]=0.60(M),[O_(2)]=0.82(M) and [SO_(3)]=1.90(M)`? `2SO_(2)(g)+O_(2)(g)hArr 2SO_(3)(g)`.

Explanation

For the given reaction equilibrium: `K_(c)=([SO_(3)]^(2))/([SO_(2)]^(2)[O_(2)])`
Given: `[SO_(2)]=0.60(M),[O_(2)]=0.82(M) and [SO_(3)]=1.90(M)`
`therefore K_(c)=((1.9)^(2)L*mol^(-1))/((0.6)^(2)xx0.82)=12.22L*mol^(-1)`.