Suleiman N.

asked • 05/14/23

If instead 0.0100 M of W, 0.0150 M of X, 0.0250 M of Y, and 0.000515 M of Z are mixed, determine the equilibrium concentrations of all species (Keq still equals 4.43 X 10-7)

If instead 0.0100 M of W, 0.0150 M of X, 0.0250 M of Y, and 0.000515 M of Z are mixed, determine the equilibrium concentrations of all species (Keq still equals 4.43 X 10-7). (assumption is needed and initially not valid).


2W (aq) + X (aq) ⇌ 2Y (aq) + Z (aq)


I believe this question requires a Double ICE Table (ICICE Table) but I am not sure. Also could you please explain which parts of the reaction can be simplified to 0 and why?



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J.R. S. answered • 05/14/23

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What is meant by "if instead..."? Is this problem stating that the initial conditions are

0.01 M W, 0.0150 M X, 0.0250 M Y and 0.000515 M Z? If so, then we proceed as follows:

2W + X <==> 2Y + Z

0.01...0.015.........0.025....0.000515....Initial

+2x....+x.............-2x........-x..................Change

0.01+2x..0.015+x..0.025-2x...0.00515-x...Equilibrium

K = [Y]2[Z] / [W]2[X] = 4.43x10-7

4.43x10-7 = (0.025-2x)2(0.000515-x) / (0.01+2x)2(0.015+x)

Solve for x and then return to the Equilibrium line of the ICE table and calculate concentrations of each species.

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Suleiman N.

This problem is from a set of practice problems that all share the same format as the previous questions. I think they meant "What happens if we change up the concentration #'s but keep the overall question the same" because in both questions the Kc is the same.
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05/14/23

Suleiman N.

After setting up the equilibrium expression, could you explain which x's can be assumed to be 0. This part confuses me a lot.
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05/14/23

J.R. S.

tutor
Well, not knowing the overall question, it's difficult to understand what is being asked here. The only x's that would be 0 in an ICE table are those that are NOT present initially. The way I read the question, all species are present initially, but that may not be the case in the original question, which we haven't seen.
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05/14/23

J.R. S.

tutor
Because the K value is so small, the reaction will go in the reverse direction.
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05/14/23

Suleiman N.

Sorry for the confusion. The first question is not needed to answer this one. I think the teacher just reused the same question but changed the numbers. Both questions ask for the same thing "determine the equilibrium concentrations of all species". For the x's being zero that i am referring to are not in the ice table. I am referring to the variables in the equilibrium expression that can be assumed to be really small so we replace them with 0 to get rid of them and solve without having to use the quadratic equation.
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05/14/23

J.R. S.

tutor
Oh, now I see. For x to be assumed to be 0, it would have to be less than a given percent of the value you are adding it to or subtracting it from. The general rule is 5%. So, go ahead and assume x is small and then solve. If it turns out to be greater than 5%, then you need to go back and use the quadratic. Because K is so small (~10^-7), x will most likely be less than 5%. Hope this helps.
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05/14/23

Suleiman N.

I was given an answer key along with the question but it doesn't go through it step by step. Is it possible that you could go through the whole question so I can see where my error is? The Correct answers are [Z] = 2.39 X 10-9 M, [Y] = 0.0240 M, [X] = 0.0255 M,[W] = 0.0110 M (assumption is needed and initially not valid).
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05/14/23

J.R. S.

tutor
Sorry, but I don't think I can help you right now. I'm on my way out and won't return until later today. From first glance, the answers don't make sense because the reactants don't increase by the same x value and the products don't decrease by the predicted values. And I'm confused by the statement "assumption is needed and initially not valid". Maybe someone else will weigh in, and if not, maybe I can look at it again later today.
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05/14/23

Suleiman N.

Ok thank you!
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05/14/23

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