At `450 K`, `K_(p) = 2.0 xx 10^(10)//"bar"` for the given reaction at equilibrium
`2SO_(2)(g) + O_(2)(g) hArr 2SO_(2)(g)`
What is `K_(c)` at this temperature?

Explanation

For the given reactions.
` " " Delta n = 2- (2+1) =-1`
` " " K_p=K_c (RT)^(Delta n )`
or ` " " K_c =(K_p)/((RT)^(Delta n )) `
or ` " " =(2.0 xx 10 ^(10) )/((0.083xx 450 )^(1)) L " mol" ^(-1)`
` " " (because R_(10 ) = 0.0831 " L bar " K^(-1) " mol"^(-1))`
` " " =2.0 xx 10 ^(10) xx 0.083 1 xx 450 L " mol "" ^(-1) `
` " " 7.48 xx 10 ^(11) " L mol"^(-1)`