Posted by **Anna** on Sunday, March 16, 2014 at 3:42pm.

This is a problem I got on my Mastering Chemistry homework:

Consider the following reaction:

H2(g)+I2(g)⇌2HI(g)

A reaction mixture at equilibrium at 175 K contains PH2=0.958atm, PI2=0.877atm, and PHI=0.020atm. A second reaction mixture, also at 175 K, contains PH2=PI2= 0.623atm , and PHI= 0.110atm

The question is: If the second reaction is not at equilibrium, what will be the partial pressure of HI when the reaction reaches equilibrium at 175 K?

I did the ICE table and actually got to this part: (0.110-2x)^2 /(0.623+x)^2 = 4.76x10-4

But, I can't seem to get the right answer. I know I have to solve for x and I got 0.492. And then I plugged that value for x into the equilibrium equation for HI which is 0.110-2x. Maybe my math is wrong but I'm not sure. Can someone please help me?

- Chemistry -
**DrBob222**, Sunday, March 16, 2014 at 4:15pm
I think it's your math. I obtained the same equation as you and it looks ok to me. I went through the math and obtained two values for x of 0.0631 (which can't be right since 0.110-2x gives a negative number) and 0.0471. That value plugged back into Kp expression gives Kp = 5.6E-4. I think if I carried the work out another place it would be closer to Kp but I didn't try that. If you wish to post your math work I'll be happy to look for the error. Your chemistry part look good to me.

- Chemistry -
**Anna**, Sunday, March 16, 2014 at 4:34pm
Thank you so much. I figured out what I did wrong.

- Chemistry -
**Victoria Miceli**, Sunday, October 16, 2016 at 11:26pm
I have this problem too but with different numbers for the second mixture. I have no idea where to go from here. If you could help that would be awesome.

Mixture 2: pH2=pI2=0.610 atm

pHI=0.107atm

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