Homework 5J.1

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Lindsey Chheng 1E
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Joined: Fri Aug 30, 2019 12:16 am

Homework 5J.1

Postby Lindsey Chheng 1E » Tue Jan 14, 2020 8:51 pm

Consider the equilibrium CO(g) + H2O(g) ⇌ CO2(g) + H2(g).
a) If the partial pressure of CO2 is increased, what happens to the partial pressure of H2?

I think that the partial pressure of H2 will decrease from the reaction shifting left to re-establish equilibrium. Is this correct?

Brooke Yasuda 2J
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Re: Homework 5J.1

Postby Brooke Yasuda 2J » Tue Jan 14, 2020 8:55 pm

Yes. If the partial pressure of CO2 increases, then Q is going to be greater than K. As a result, the reverse reaction is going to be favored, and the partial pressure of H2 decreases.

Ariel Davydov 1C
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Re: Homework 5J.1

Postby Ariel Davydov 1C » Tue Jan 14, 2020 8:56 pm

When dealing with problems with shifts in equilibrium, try to approach problems with partial pressures as you would problems with molar concentrations. If you set up a K equation in terms of partial pressures, you will see that if you increase the partial pressure of a product, the system will shift to the left and increase the partial pressures of the reactants in order to maintain the same produce/reactant ratio at equilibrium. This is consistent with Le Chatelier’s principle, which states that a reaction will shift in order to minimize the stress of a change. Thus, if we increase the partial pressure of CO, the partial pressure of H2 will decrease in order to maintain the same Kp at equilibrium.

Lindsey Chheng 1E
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Joined: Fri Aug 30, 2019 12:16 am

Re: Homework 5J.1

Postby Lindsey Chheng 1E » Tue Jan 14, 2020 9:03 pm

thank you!

Caroline Beecher 2H
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Re: Homework 5J.1

Postby Caroline Beecher 2H » Wed Jan 15, 2020 9:13 pm

I was also curious about this question! Applying Le Chatelier's principle makes sense also for part B of this question. In part B, if the partial pressure of CO is decreased, then the partial pressure of CO2 will decrease because the reverse reaction will be favored and there will be less products.


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