# Chemistry

posted by Anna

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?

1. DrBob222

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.

2. Anna

Thank you so much. I figured out what I did wrong.

3. Victoria Miceli

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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