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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

`K_p` and `K_c` are related as :
`K_p=K_c(RT)^(Deltan_g)` or `K_c=K_p/(RT)^(Deltan_g)`
T=700 K , R=8.31 J `mol^(-1)K^(-1) , Deltan_g` =(2+1) -2=1
`K_p=1.8xx10^(-3)` kPa = 1.8 Pa
`K_c=(1.8 Pa)/[(8.31 J mol^(-1) K^(-1))(700K)]^(1)`
`=(1.8 Nm^(-2))/([(8.31 N "mol"^(-1)K^(-1))(700 K)])` (`"because" Pa = N m^(-2)`, J= N m)
`=3.09xx10^(-4) "mol m"^(-3)` or `3.09xx10^(-7) "mol dm"^(-3)` or
`3.09xx10^(-4) mol//l`
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