Textbook Problem 5.57

[Elsie_Lin_2K]

The question is asking about the amount of NO that must be added at equilibrium. After I set up an ICE table, I found that the value of x is 0.24. However, when I plugged in the values 6.0*10^3=[0.24^2]/[0.005*NO], I would get 0.00192 instead of 0.242 moles which is the answer from the book.

[Joanna Zhao 1J]

That's because the answer is looking for how much NO was added at the beginning, aka the INITIAL concentration of NO. If the equilibrium/final concentration was 0.00192 M (your calculator is correct), and NO lost 0.240 M in the process, then initial concentration of NO must be 0.00192 M + 0.240 M = 0.24192 or 0.242 M.

[WendyN_L2]

As the previous post said, we are looking for the initial concentration of NO before 0.240 mols of it was used up. So when setting up your ICE table for NO it should be (x-0.240), x being the initial amount of NO, and adding that into the equation
6x10^3=(.240)^2/(x-0.24)(.005) and you should get the answer. Hope that helps

[905767718]

I think the issue here is how you're solving for x in the chemical equilibrium equation. Your equation should be K = 6000 = (0.240)^2/((0.005)(x-0.240)). From there you just solve for x using algebra. The main reason that this equation is a little different is because you're solving for the initial concentration of NO compared to when you solve for the equilibrium concentration of a molecule.