Answer to Question #128673 in General Chemistry for A

Question

Good day,
where R=8.314 L⋅kPa/(K⋅mol)In Kp=Kc(RT)Δn
For the reaction
X(g)+3Y(g)⇌3Z(g)
Kp = 1.24×10−5 at a temperature of 169 ∘C
Calculate the value of Kc.I calculated to be 4.49x10^-4 but says incorrect.Please help

Thanks

Answer 1

For this reaction Δn is 3-(1+3)=-1 (Δn is the sum of the stoichiometric coefficients of the 

 gaseous products minus the sum of the stoichiometric coefficients of the gaseous reactants).

Thus, Kc=Kp/(RT)^(Δn) (temperature in Kelvins)

Kc=1.24*10^-5/(8.314*(169+273.15))^(-1)=0.0456