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?
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.
Thank you so much. I figured out what I did wrong.
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
Sure, I can help you with that question.
To solve this problem, you correctly set up the ICE table and arrived at the equation:
(0.110 - 2x)^2 / (0.623 + x)^2 = 4.76 x 10^-4
This equation represents the equilibrium expression for the reaction, where x represents the change in the concentration of HI. By solving for x, you can determine the change in partial pressure of HI when the reaction reaches equilibrium.
To solve the equation, follow these steps:
1. Expand the equation to eliminate the square terms:
(0.110 - 2x)^2 = 4.76 x 10^-4 * (0.623 + x)^2
2. Expand and simplify the terms on both sides of the equation:
0.0121 - 0.44x + 4x^2 = 4.76 x 10^-4 * (0.388129 + 1.246x + x^2)
3. Distribute the 4.76 x 10^-4 to the terms on the right side:
0.0121 - 0.44x + 4x^2 = 1.84622 x 10^-4 + 5.91904 x 10^-4x + 4.76 x 10^-4x^2
4. Collect like terms on both sides and rearrange the equation:
3.50378 x 10^-4x^2 - 0.44x + 0.01025378 = 0
5. Rearrange the equation to be in standard quadratic form:
3.50378 x 10^-4x^2 - 0.44x + 0.01025378 = 0
6. Now, you can solve the quadratic equation using any method you prefer. One common approach is to use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, the coefficients are:
a = 3.50378 x 10^-4
b = -0.44
c = 0.01025378
Plugging these values into the quadratic formula will give you the two possible solutions for x. Note that one of the solutions will likely be extraneous, so it's important to check it.
Once you find the correct value for x, you can substitute it into the equation for HI to obtain the partial pressure of HI at equilibrium.
I hope this explanation helps you in solving the problem. Let me know if you have any further questions!