#26 on post-assessment

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005199302
Posts: 108
Joined: Fri Sep 28, 2018 12:15 am

#26 on post-assessment

Postby 005199302 » Wed Jan 09, 2019 7:51 pm

I don't know how to do the following question. Does anyone know how and can explain it conceptually?

Question: 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).

Kim Tran 1J
Posts: 62
Joined: Fri Sep 28, 2018 12:24 am

Re: #26 on post-assessment

Postby Kim Tran 1J » Wed Jan 09, 2019 9:13 pm

You can use the I.C.E. box to help visualize/solve this.

Initial molar concentration (I)
Change in molar concentration (C)
Equilibrium molar concentration (E)

N2 H2 NH3
I 0.250 0.500 0
C -0.075 -0.225 +0.15
E 0.175 0.275 0.15

Given the change in the concentration of NH3, you can find the change in N2. The ratio of N2 to NH3 is 1:2, so there is half the amount of N2 for every one mole of NH3. So if the concentration of NH3 is increased by 0.15mol/L, then the N2 is decreased by 0.075mol/L, which is half of 0.15 (0.15 / 2 = 0.075). So the concentration of N2 at equilibrium is the initial minus the change which would be (0.250 - 0.075 = 0.175mol/L) and would be rounded to 0.18M considering sig figs.


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