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At 700 K, the equilibrium constant K(p) ...

At `700 K`, the equilibrium constant `K_(p)` for the reaction
`2SO_(3)(g)hArr2SO_(2)(g)+O_(2)(g)`
is `1.80xx10^(-3) kPa`. What is the numerical value of `K_(c )` in moles per litre for this reaction at the same temperature?

Text Solution

Verified by Experts

Here `n_(p) =3 " moles" , n_(r) = 2 " moles "`
`:. " "Deltan = n_(p) - n _(r) = 3- 2 = 1 `
` K_(p) = 1.80 xx 10^(-3) kPA = 1. 80 PA = (1.80)/(10^(5)) "bar" = 1.80 xx 106(-5) " bar"`
`R= 0.0831 L " bar " K^(-1) mol^(-1), T=700 K`
Using the relation , `K_(p) = K_(c) (RT)^Delta n`
`K_(c) = (K_(p))/(RT)= (1.80 xx 10^(-5) "bar")/(0.0831 L " bar " K^(-1) xx 700 K )= 3.09 xx 106(-7) " mol " L^(-1)`
`" Alternatively, " K_(c) = (K_(p))/(RT) =(1.8Pa)/((8.314 JK^(-1) mol^(-1))(700 K))" "( :' Pa = N m^(-2), J = N m)`
`= 3.09 xx 10^(-4)" mol " m^(-3) = 3.09 xx 10^(-7)" mol " dm^(-3) or" mol " L^(-1)`
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