### 5H.1 part B

Posted: **Wed Jan 09, 2019 8:26 pm**

by **Emma Randolph 1J**

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?

### Re: 5H.1 part B

Posted: **Wed Jan 09, 2019 8:37 pm**

by **Ray Huang 1G**

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.

### Re: 5H.1 part B

Posted: **Wed Jan 09, 2019 8:46 pm**

by **KimGiang2F**

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.