5H.1 part B

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Emma Randolph 1J
Posts: 65
Joined: Fri Sep 28, 2018 12:29 am

5H.1 part B

Postby Emma Randolph 1J » Wed Jan 09, 2019 8:26 pm

For the reaction N2(g) + 3H2(g) -> 2 NH3(g) at 400. K, K = 41. Find the value of K for each of the following reactions at the same temperature: (a) 2 NH3(g) -> N2(g)+3 H2(g) (b) 1/2 N2(g) + 3/2 H2(g) ->NH3(g) (c) 2 N2(g)+6 H2(g) -> 4 NH3(g). I understand that for part a you just put 1/41 to solve for K, but I don't understand how to solve for K for parts b and c, could someone explain this to me?

Ray Huang 1G
Posts: 30
Joined: Fri Sep 28, 2018 12:20 am

Re: 5H.1 part B

Postby Ray Huang 1G » Wed Jan 09, 2019 8:37 pm

For part b the equation is what would happen if you took the original and divide it by two. Part c is when you take the original and multiply by two. I don't think we have yet to go over how to manipulate k when you do a scalar multiplication to the reaction.
Last edited by Ray Huang 1G on Wed Jan 09, 2019 9:13 pm, edited 1 time in total.

KimGiang2F
Posts: 31
Joined: Fri Sep 28, 2018 12:19 am

Re: 5H.1 part B

Postby KimGiang2F » Wed Jan 09, 2019 8:46 pm

In reaction presented, N2(g) + 3H2(g) -> 2 NH3(g) at 400, K = 41. The reaction of part B, 1/2 N2(g) + 3/2 H2(g) ->NH3(g), changed in the amount of moles. However, the ratio of the moles of the equation of part B is proportional to the ratio of the moles of the equation of part A by one-half. We assume that there is no change in temperature so the equilibrium must be constant. Essentially, since the ratio of reactant to product remains the same, the equilibrium must also be the same. Therefore, K=41. Similarly, the ratio of the moles of the equation of part C, 2 N2(g)+6 H2(g) -> 4 NH3(g), is proportional to the ratio of the moles of the equation of part A by two times the amount. Once again, the ratio of reactant to product remains the same, therefore, K=41.


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