Winter 2013 Midterm: Question 6A  [ENDORSED]


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Karla_Coronado_1J
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Winter 2013 Midterm: Question 6A

Postby Karla_Coronado_1J » Sat Feb 11, 2017 10:26 am

A gaseous mixture consisting of 2.23 mmol N2 and 6.69 mmol H2 in a 500.-mL container was heated to 600. K and allowed to reach equilibrium. Will more ammonia be formed if that equilibrium mixture is then heated to 700. K? Give a brief reason and answer.
For N2(g) + 3H2(g) <---> 2NH3(g), K = 1.7 x 10-3 at 600. K and 7.8 x 10-5 at 700. K.

May someone explain to me how I go about answering this question?

Thanks in advance.

Amy_Bugwadia_3I
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Re: Winter 2013 Midterm: Question 6A  [ENDORSED]

Postby Amy_Bugwadia_3I » Sat Feb 11, 2017 11:00 am

We are given the equilibrium constant K for both the temperatures. Since it's 1.7 x 10^(-3) at 600K but 7.8 x 10^(-5) at 700K, we know that K is smaller at the higher temperature. When we increase the temperature of ammonia (NH3, on the product side), the reaction will form more reactants due to Le Chatlier's principle. Favoring the formation of the reactants, by necessity, means that the amount of product (in this case, NH3) will decrease. Thus, we will have less amonia after heating to 700K.


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