Question

Good day,please help me please.
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
Thank you very much.

Answer 1

Solution.

$Kp = Kc \times (RT)^{\Delta n}$

$\Delta n = 3 - (1+3) = -1$

$T(K) = T(^oC) + 273$

T(K) = 442 K

$Kc = \frac{Kp}{(RT)^{\Delta n}}$

$Kc = 4.56 \times 10^{-2}$

Answer:

$Kc = 4.56 \times 10^{-2}$