Question

At a given temperature, 3.12 atm of H2 and 5.52 atm of I2 are mixed and allowed to come to equilibrium. The equilibrium pressure of HI is found to be 1.738 atm. Calculate Kp for the reaction at this temperature.
H2(g) + I2(g) <=> 2 HI(g). Give your answer to 3 decimal places.

Answer 1

Answer: 0.596

Explanation:

For this problem, we want to find $K_{p}$. To do so, we will need to use the ICE chart. The I in ICE is initial quantity. In this case, it is the initial pressure. Pressure is in atm. The C in ICE is change in each quantity. The E is equilibrium.

          H₂(g) + I₂(g) ⇄ 2HI(g)

I          3.12     5.52         0

C      -0.869  -0.869   +1.738

E       2.251    2.251     1.738

For the steps below, refer to the ICE chart above.

1. Since we were given the initial of H₂, I₂ and equilibrium of 2HI, we can fill those into the chart.

2. Since we were not given the initial for 2HI, we will put 0 in their place.

3. For the change, we need to add pressure to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x. We know that the equilibrium of 2HI is 1.738, we know the change is 0.869 because 1.738/2=0.869. Since there are 2 moles of HI, we must divide the equilibrium by 2 to find x, so that we can fill that into the reactants side.

4. With the equilibrium values, we can find the equilibrium pressure. It is products over reactants. We use the values of E.

$K_{p} =\frac{[HI]^2}{[H_{2}][I_{2} ] }$

The [HI]² comes from 2HI. We more the moles to the exponent when we are calculating the equilibrium pressure.

$K_{p} =\frac{(1.738)^2}{(2.251)(2.251)} =0.596$

We typically don't put units because K is unitless, but know that the units is atm throughout the entire problem.