An empty steel container is filled with 0.0590 atm of HF. The system is allowed to reach equilibrium. If Kp = 2.76 for the reaction below, what is the equilibrium partil pressure of H2?

Question

2HF (g) = H2(g) + F2(g)

Answer 1

2HF (g) = H2(g) + F2(g)

0.059........0..........0........Initial

-2x..........+x.........+x.......Change

0.059-2x...x..........x........Equilibrium

Kp = (H2)(F2) / (HF)² = (x)(x) / (0.059-2x)²

2.76 = x² / 4x² - 0.236x + 3.4810^-3

10.04x² - 0.65x + 9.6x10^-3 = 0

x = 0.023 atm = equilibrium partial pressure of H2

(be sure to check the math)