The question is asking if products or reactants will be favored by an increase of total pressure.
E) 2HD(g)+H2(g) -> D2(g)
The solution manual says no change because there is the same number of moles on each side of the equation. Can someone explain this? I thought it would shift towards the products.
5J.5 D
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Re: 5J.5 D
I think you meant to write 2HD (g)--> H2 (g)+ D2 (g)
When you increase the pressure of the system, you expect the reaction to favor the side with the less moles of gas to relieve the pressure (Le Chatelier's principle). However, in this case, there are 2 moles of gas in the reactants and 2 moles of gas in the products side. So, the reaction wouldn't shift to either side.
When you increase the pressure of the system, you expect the reaction to favor the side with the less moles of gas to relieve the pressure (Le Chatelier's principle). However, in this case, there are 2 moles of gas in the reactants and 2 moles of gas in the products side. So, the reaction wouldn't shift to either side.
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Re: 5J.5 D
Faaizah Arshad 1H wrote:I think you meant to write 2HD (g)--> H2 (g)+ D2 (g)
When you increase the pressure of the system, you expect the reaction to favor the side with the less moles of gas to relieve the pressure (Le Chatelier's principle). However, in this case, there are 2 moles of gas in the reactants and 2 moles of gas in the products side. So, the reaction wouldn't shift to either side.
Are you saying that because of the moles the reaction is at equilibrium?
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Re: 5J.5 D
Yes, I believe so. It is important to consider moles, but since they are equal on either side (2), then the reaction wouldn't shift.
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