The equilibrium constant, `K_(p)` for the reaction
`2SO_(2)(g)+O_(2)(g)hArr 2SO_(3)(g)`
is `4.0 atm^(-1)` at 1000 K. What would be the partial pressure of `SO_(2)` if at equilibrium the amount of `SO_(2)` and `SO_(3)` is the same ?

Options

A. 16.0 atm
B. 0.25 atm
C. 1 atm
D. 0.75 atm

Answer

B

Explanation

Equal amounts of `SO_(2)` and `SO_(3)` would mean that they would have the same partial pressures.
`K_(p)=(p(SO_(3))^(2))/(p(SO_(2))^(2)xx p(O_(2)))`
Since `p(SO_(3))=p(SO)_(2)`
`therefore K_(p)=(1)/(p(O_(2)))`
or `p(O_(2))=(1)/(K_(p))=(1)/(4)=0.25` atm