Chem. Equilibrium Part 2 Post-Module Assess. #26  [ENDORSED]

Moderators: Chem_Mod, Chem_Admin

Grace Lee 3G
Posts: 19
Joined: Fri Sep 29, 2017 7:07 am

Chem. Equilibrium Part 2 Post-Module Assess. #26

Postby Grace Lee 3G » Wed Nov 29, 2017 11:00 pm

26. A mixture initially consisting of 0.250 N2 (g) and 0.500 M H2 (g) reacts to form NH3 (g) which is 0.15 M NH3 (g) at equilibrium. Calculate the concentration of N2 (g) at equilibrium for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g).

A. 0.18 M

B. 0.10 M

C. 0.33 M

D. 0.40 M

Aijun Zhang 1D
Posts: 53
Joined: Tue Oct 10, 2017 7:13 am

Re: Chem. Equilibrium Part 2 Post-Module Assess. #26

Postby Aijun Zhang 1D » Thu Nov 30, 2017 12:01 am

I think the answer is B.

N2(g) + 3H2(g) ⇌ 2NH3(g).
I 0.250 0.500 0
C -0.075 -0.225 +0.15
E 0.175 0.275 0.15

Basically, the change of concentration of NH3 indicates the change of N2. If concentration of NH3 is increased by 0.15mol/L, then the N2 is decreased by 0.075mol/L, calculated by the coefficients(0.15/2*3). So the concentration of N2 at equilibrium is 0.175mol/L, approximated to 0.18M.

isauramora3K
Posts: 20
Joined: Fri Sep 29, 2017 7:07 am

Re: Chem. Equilibrium Part 2 Post-Module Assess. #26  [ENDORSED]

Postby isauramora3K » Sat Dec 02, 2017 1:47 pm

The correct answer is 0.175 or 0.18 as stated above, so the answer is A not B as stated above by mistake


Return to “Non-Equilibrium Conditions & The Reaction Quotient”

Who is online

Users browsing this forum: No registered users and 1 guest