Am I correct on setting up and solving the following question?... 2H2O2(g) <=> 2

Question

Am I correct on setting up and solving the following question?...

2H2O2(g) <=> 2H2O(g) + O2(g) ...1.75 moles of H2O2 were placed in a 2.5L reaction chamber at 307 degrees C. After equiibrium was reached, 1.20 moles of H2O2 remained. What is Kc for the reaction?

I did... 1.75/2.5 moles of H2O2 = 0.48 Molarity, then I assumed 0.48 Molarity for H2O and 0.24 Molarity O2 in order to keep each balanced with one another.

Then plug the numbers into the Kc equation: [(0.48)^2 x (0.24)] / (0.48)^2 = 0.24 ...but I don't think I have the correct answer. Any help?

Explanation / Answer

2H2O2 <----------> 2H2O + O2

Kc = {[H2O]2*[O2]}/[H2O2]2

Now, initially, [H2O2] = 1.75/2.5 = 0.7 M & [H2O] = [O2] = 0 M

Let at eqb. [H2O2] = (0.7 - 2x) M & [H2O] = 2x M & [O2] = x M

Now, given that at eqb. [H2O2] = 1.2/2.5 = 0.48 M

Hence x = 0.11

Thus, at eqb, [H2O] = 2x = 0.22 M & [O2] = 0.11 M

Thus, Kc = {(0.22)2*0.11}/(0.48)2 = 0.023

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