Chemistry 2 - Equilibrium Constant

Question

At 423 K, this reaction has a 𝐾c value of 0.0659.

X(g)+3Y(g)↽−−⇀2Z(g)

Calculate 𝐾p at 423 K. Note that the pressure is in units of atmosphere (atm).

𝐾p= ?

Answer 1

Kp = Kc(RT)^∆n

Kp = ?

Kc = 0.0659

R = gas constant = 0.0821 Latm/Kmol

T = temp in Kelvin = 423K

∆n = change in moles of gas = 2 - 4 = -2

Solving for Kp we have...

Kp = 0.0659 (0.0821*423)^-2 = 0.0659 x 0.1697

Kp = 0.0112

(be sure to check the math)

Answer 2

This question is testing the relationship between Kp and Kc, which is given by thee equation

Kp =Kc (RT)^delta n

where delta n = the diference between the sum of the product coefficients minus the sum of the reactant coefficients

Kc is given as 0.0659

R=0.08206 L atm/mok K

T=423 degrees Kelvin

delta n = 2 -5 = -2

so plugging in should get

Kp = 0.0659 [0.08206 * 423] ^ -2

or 0.0659 [34.71138] ^ -2 = 0.0659 x [8.29958 x 10 E4] = 5.469 x 10^-5