5J.3

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ayushibanerjee06
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5J.3

Postby ayushibanerjee06 » Mon Jan 13, 2020 2:59 pm

The question is: The four gases NH3, O2, NO, and H2O are mixed in a reaction vessel and allowed to reach equilibrium in the reaction 4 NH3(g) + 5 O2(g) <-> 4 NO(g) + 6 H2O(g). Why is there less NH3 when NO is removed?

Benjamin Feng 1B
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Re: 5J.3

Postby Benjamin Feng 1B » Mon Jan 13, 2020 3:10 pm

When you remove NO, the system wants to balance out agian and produce more. This means that the forward reaction proceeds and NH3 is used up.

Christineg1G
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Re: 5J.3

Postby Christineg1G » Mon Jan 13, 2020 3:58 pm

According to Le Chatelier's principle, removing NO will lead to the formation of more products. So to maintain equilibrium, the amount of NH3 will decrease. Hope this helps!

anjali41
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Re: 5J.3

Postby anjali41 » Mon Jan 13, 2020 4:04 pm

When NO is removed, the product of the forward reaction is being decreased. In order to reach equilibrium again, the forward reaction will proceed and use the reactants to make more product. So, the reactants, including NH3, decrease as a result.

Charisse Vu 1H
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Re: 5J.3

Postby Charisse Vu 1H » Mon Jan 13, 2020 4:05 pm

When you remove NO, the reaction favors the products because it wants to reach equilibrium. Since there is now less NO, and therefore products, the reaction proceeds forward to produce more products. This causes NH3 (reactant) to decrease.

Simon Dionson 4I
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Re: 5J.3

Postby Simon Dionson 4I » Tue Jan 14, 2020 6:57 pm

Removal of a product will cause the equilibrium to shift to the right, meaning there will be more reactants consumed to reach equilibrium again.


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