Exercise 1. The following reaction has Kc = 1.8 x 10-8 at 45.6 oC. What is Kp?
A (g) + 2 B (g) ⇄ AB2 (g)
Kp = Kc(RT)Δn
Δn = 1 mol – 3 mol = -2. K = 45.6 oC + 273.15 = 318.75 K
Kp = Kc(RT)Δn = 1.8 x 10-8 (0.0821 (L⋅atm)/(mol⋅K) x 318.75)-2 = 2.6 x 10-11
Back to Relationship Between Kc and Kp
Exercise 2. Calculate Kc for the following reaction.
N2 (g) + 2 H2 (g) ⇄ N2H4 (g) Kp = 4.09 x 10-29 at 45.5oC.
\(\displaystyle K_c\;=\;\frac{K_p}{(RT)^{Δn}}\)
Δn = -2
K = 45.5 + 273.15 = 318.65 K
\(\displaystyle K_c\;=\;\frac{4.09\times 10^{-29}}{(0.0821\frac{L⋅atm}{mol⋅K}\times 318.65\;K)^{-2}}\;=\;\mathbf{2.80 \times 10^{-26}}\)