11.37 6th edition

Moderators: Chem_Mod, Chem_Admin

CaminaB_1D
Posts: 63
Joined: Fri Sep 28, 2018 12:16 am

11.37 6th edition

Postby CaminaB_1D » Tue Jan 15, 2019 4:38 pm

I don't understand how to calculate K for part c of this problem, can someone please explain the reasoning for the answer that the solution manual comes to?

11.37 For the reaction N (g) + 3 H (g) ∆ 2 NH (g) at 223
400. K, K 41. Find the value of K for each of the following reactions at the same temperature:

(c) 2 N2(g) + 6 H2(g) ∆ 4 NH3(g)

Kate Chow 4H
Posts: 61
Joined: Fri Sep 28, 2018 12:28 am

Re: 11.37 6th edition

Postby Kate Chow 4H » Tue Jan 15, 2019 4:45 pm

You're given that k = 41 when N2+3H2 --> 2NH3. In part c of the question, the reaction is now 2N2 + 6H2 --> 4NH3. Notice that all of the coefficients were simply multiplied by 2 in part c. If you calculate K using products/reactants with the coefficients as the powers, you realize that the coefficients are doubled, so the powers are squared. Overall, for part c, k = 41^2.

kamalkolluri
Posts: 65
Joined: Fri Sep 28, 2018 12:17 am
Been upvoted: 1 time

Re: 11.37 6th edition

Postby kamalkolluri » Tue Jan 15, 2019 4:45 pm

When you multiply a certain reaction by a number n, the new K for the new equation is the old K to the power of n. So here, the equation is multiplied by 2 so the K of the reaction would be K^2 so 41^2.

204929947
Posts: 76
Joined: Fri Apr 06, 2018 11:03 am

Re: 11.37 6th edition

Postby 204929947 » Fri Feb 08, 2019 8:02 pm

I understand part C but can someone please explain part A and B


Return to “Equilibrium Constants & Calculating Concentrations”

Who is online

Users browsing this forum: No registered users and 2 guests