Equilibrium and using ICE tables

Question

An empty steel container is filled with 0.0400 atm of HF. The system is allowed to reach equilibrium. If Kp = 2.76 for the reaction below, what is the equilibrium partial pressure of H₂? 2 HF (g) ⇌ H₂ (g) + F₂ (g)

Answer 1

So, Cali M. You've posted 4 consecutive problem all seemingly dealing with equilibrium and ICE tables. I will do this problem, and I'll look at the others, but hopefully after reviewing the info here, you'll be able to do the others. Have you even tried to do these? Where is it that you are having a problem?

We begin by writing the balanced equation:

2HF(g) <==> H₂(g) + F₂(g)

We then set up an ICE table. I is for initial; C is for change; E is for equilibrium

2HF(g) <==> H₂(g) + F₂(g)

0.0400...........0..........0..........Initial

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

0.04-2x...........x..........x..........Equilibrium

Next, we write the Kp expression:

Kp = 2.76 = (P_H2)(P_F2) / (P_HF)²

2.76 = (x)(x) / (0.04-2x)²

2.76 = x² / 4x² - 0.16x + 0.0016

x² = 11.04x² - 0.4416x + 0.004416

10.04x² - 0.4416x + 0.004416 = 0

x = 0.0154 atm (be sure to check the math)

Use this value of x and the value in the ICE table for H2 to get the equilibrium pressure of H₂:

P_H2 = x = 0.0154 atm