Question

At 450 K, K_p= 2.0 × 10¹⁰/bar for the given reaction at equilibrium.

\[\ce{2SO2(g) + O2(g) ⇌2SO3 (g)}\]

What is K_c at this temperature?

Answer 1

For the given reaction,

Δn = 2 – 3 = – 1

T = 450 K

R = 0.0831 bar L bar K^–1 mol^–1

`"K"_"p"` = 2.0 × 10¹⁰ bar ^–1

We know that,

`"K"_"P" = "K"_"C"("RT")triangle"n"`

`= 2.0 xx 10^10 "bar"^(-1) = "K"_"C"(0.00831" L bar K"^(-1) "mol"^(-1) xx 450" K")^(-1)`

`=> "K"_"C" = (2.0 xx 10^10 "bar"^(-1))/(0.0831 " L bar K"^(-1) "mol"^(-1)xx 450" K")`

`= (2.0 xx 10^(10) "bar"^(-1))(0.0831 "bar K"^(-1) "mol"^(-1) xx 450" K")`

`= 74.79 xx 10^10 "L mol"^(-1)`

`= 7.48 xx 10^(11) "L mol"^(-1)`

`= 7.48 xx 10^(11) " M"^(-1)`