Answer to Question #128592 in General Chemistry for A

Question

Good day,

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

Thanks
.

Answer 1

K_p = K_c×(R×T)^Δn

K_c = K_p/(R×T)^Δn

There are 3 mol of gas on the product side of the chemical equation, and there are 1+3=4 mol of gaseous reactants, so Δn = (3-4) = -1 in this case.

T = 169 + 273.15 = 442.15 K

R = 0.08206 L×atm/mol×K

K_c = 1.24×10^-5/(0.08206×442.15)^-1 = 1.24×10^-5/(0.08206×442.15)^-1 = 1.24×10^-5/(36.28)^-1 = 36.28×1.24×10^-5 = 44.98×10^-5 = 4.49×10^-4

Answer: 4.49×10^-4