Given that the equilibrium constant for the reaction,
`2SO_(2(g)) +O_(2(g)) Leftrightarrow 2SO_(3(g))` has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature?
`SO_(3(g)) Leftrightarrow SO_(2(g))+1/2 O_(2(g))?`

Options

A. `1.8 xx 10^(-3)`
B. `3.6 xx 10^(-3)`
C. `6.0 xx 10^(-2)`
D. `1.3 xx 10^(-5)`

Answer

C

Explanation

`2SO_(2(g)) +O_(2(g)) Leftrightarrow 2SO_(3(g)) …………(i)`
K=278
By reversing the equation (i), we get
`2SO_(3(g)) Leftrightarrow 2SO_(2(g))+O_(2(g)) ……………(ii)`
Equilibrium constant for this reaction is `K.=1/K=1/(278)`
By dividing the equation (ii) by 2, we get desired equation
`SO_(3(g)) Leftrightarrow SO_(2(g))+1/2 O_(2(g))` ...........(iii)
Equilibrium constant for this reaction
`K..=sqrt(K.)=sqrt(1/K)=sqrt(1/(278))=0.0599 approx 0.06 or 6 xx 10^(-2)`