what will the pressure of each of these gases be....?

Question

can someone explain this problem to me, my teacher did not get a chance to go over it.

At 698.6 K, a mixture of the three gases H2, I2, and HI is at equilibirum when PH2 = PI2 = 0.02745 atm and PHI = 0.2024 atm. If HI is added to the reaction vessel at constant tmeperature, so that PHI is suddenly increased to 0.8000 atm, what will the pressure of each of these gases be when the system returns to equilibrium?

thanks

Answer 1

We can write the reaction in whatever way we want.
Let's write it as

2HI <=> H2 + I2

Then
Kp= P(H2)*P(I2) / P(HI)^2 = 0.02745* 0.02745 / 0.2024 =0.00372

.. .. .. .. .. 2HI <=> .. H2 .. .. +.. I2
Initial .. .. 0.8.. .. .. 0.02745.. 0.02745
React .. .. 2x
Produce .. .. .. .. .. .. x .. .. .. .. .. x
At Equil. 0.8-2x .. 0.02745+x ..0.02745+x

Kp= P(H2)*P(I2) / P(HI)^2 = (0.02745+x)^2/(0.8-2x)^2 =>
Kp= [(0.02745+x) / (0.8-2x)]^2 =>
(0.02745+x) / (0.8-2x) = squareroot (Kp)=>
0.02745+x= 0.8*SQRT(Kp) -2*SQRT(Kp)x =>
x=(0.8*SQRT(0.00372) -0.02745) / (2*SQRT(0.00372)+1) =0.01902 so

PHI=0.8-2*0.01902 = 0.76196 atm
P(H2)= P(I2)= 0.02745+0.01902 = 0.04647 atm.

Answer 2

First we need to write a balanced equation for teh reaction.

H2(g) + I2 (g) ----------------> 2HI(g)

From thie we need the find Kp, the equilbrium constant, which is given by

Kp = PHI^2/(PH2*PI2)

we know the values of the partial pressure of each gas at equilbrium=

Kp = 0.2024^2/(0.02745*0.02745)

Kp = 268.6122

We can let the new equilbrium partial pressure of Hydrogen to be x, then the new partial pressure of iodine will also be x (as there are are only one mole of each).

The new equilbrium partial pressure of hydrogen iodide will be (0.8-2x). This is because two moles are formed. We can then re-calculate Kp.

Kp = ((0.8-2x)^2)/(x*x)

Kp= (0.8-2x)*(0.8-2x)/x^2

Kp = (0.64 - 1.6x -1.6x + 4x^2)/x^2

Kp = (0.64 -3.2x + 4x^2)/x^2

We know that Kp= 268.6122

268.6122x^2 =4x^2 -3.2x +0.64

264.6122x^2+ 3.2x -0.64= 0

Using the quadratic rules gives x to be 0.0435 or -0.055 (absurd as the partial pressure of hydrogen and iodine cannot be negative)

The partial pressures at equilbrium are

Hydrogen = 0.0435atm

Iodine = 0.0435 atm

Hydrogen iodide = 0.8- (2*0.0435)= 0.713atm