## 7th ed. 5G.5 decomposition??

aisteles1G
Posts: 117
Joined: Fri Sep 28, 2018 12:15 am

### 7th ed. 5G.5 decomposition??

Hello, for part C I don't understand how they are finding the equilibrium concentrations for X2, in the solutions manual they take the initial pressure and multiply it by whats left over after decomposition, why? Thanks for any help!

Julia Go 2L
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Joined: Sun Sep 30, 2018 12:17 am
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### Re: 7th ed. 5G.5 decomposition??

Partial pressure is proportional to the number of molecules. At equilibrium, 5/11 X2 molecules remain. This means that 45.56% of the molecules in the flask are X2 molecules. Using this, we can calculate the partial pressure of X2 by multiplying the initial pressure by 0.4556. We do this because the total pressure exerted by a mixture of gases is the sum of the partial pressures of component gases.

To find the partial pressure of X, we use a similar process. Multiply the initial pressure by the percentage of X molecules at equilibrium. However, since 1 mole of X2 molecules produces 2 moles of X molecules. We have to multiply the previous value by 2.

Next, just plug in your two values into the equilibrium constant expression to solve for K.