Given that equilibrium constant for the reaction `2SO_(2)(g) + O_(2)(g)hArr2SO_(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)hArrSO_(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) hArr 2SO_(2)(g)`
`K_(c) =278`
For the reverse reaction :
`2SO_(3)(g) hArr 2SO_(2)(g) +O_(2)(g),`
`K_(c)=(1)/(K_(c))=(1)/(278)`
For the reaction
`SO_(3)(g) hArr SO_(2)(g) +O_(2)(g)`
`K_(c)'' =K_(c)' = ((1)/(278))^(1//2) =0.06 =6 xx 10^(-2)`