At 2257 K and 1.00 bar total pressure, water is 1.77 percent dissociated at quil

Question

At 2257 K and 1.00 bar total pressure, water is 1.77 percent dissociated at quilibrium by way of the reaction 2H2O(g) <-- --> 2H2(g) + O2(g). Calculate K.

I know the answer is 2.86e-6 and I attempted many times to do the entire problem over and over but I can't seem to get the answer in the back of the book.

I understand where all the values come from. And I used (Ph2)^2PO2/(PH2O)^2P(standard)

and plug in all the values and tried to solve it but I couldn't get the answer. Can someone show me all the steps?

At 2257 K and 1.00 bar total pressure, water is 1.77 percent dissociated at quilibrium by way of the reaction 2H2O(g) I know the answer is 2.86e-6 and I attempted many times to do the entire problem over and over but I can't seem to get the answer in the back of the book. I understand where all the values come from. And I used and plug in all the values and tried to solve it but I couldn't get the answer. Can someone show me all the steps?

Explanation / Answer

Assume one mole of H2O in one liter of water initially

Initial concentration = 1/ 1 = 1 M

Water is 1.77 % dissociated

So amount dissociated = 0.0177

2 H2O <------> 2 H2 + O2

1 M 0 0

- 2(0.0177) +2(0.0177) + 0.0177

0.9646 0.0354 0.0177

K = [H2]^2 [O2] / [H2O]^2

= (0.0354)^2 * 0.0177 / (0.9646)^2

= 2.38 * 10^-5   

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