5I.25  [ENDORSED]

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KNguyen_1I
Posts: 86
Joined: Sat Aug 17, 2019 12:16 am

5I.25

Postby KNguyen_1I » Wed Jan 08, 2020 8:24 pm

5I.25 A reaction mixture is prepared by mixing 0.100 mol SO2,
0.200 mol NO2, 0.100 mol NO, and 0.150 mol SO3 in a reaction
vessel of volume 5.00 L. The reaction SO2(g) + NO2(g) <=>
NO(g) + SO3(g) is allowed to reach equilibrium at 460 C, when
Kc = 85.0. What is the equilibrium concentration of each
substance?'

I'm completely lost in this problem- can someone step by step walk me through what we're supposed to do?

VPatankar_2L
Posts: 84
Joined: Thu Jul 25, 2019 12:17 am
Been upvoted: 1 time

Re: 5I.25  [ENDORSED]

Postby VPatankar_2L » Wed Jan 08, 2020 9:09 pm

1. First, calculate the concentration of all gasses by dividing the amount in mols (this is given) by the volume (5L)
2. Set up your ICE box (as shown below) y writing down the initial concentration for all gases
3. Write out the equilibrium constant expression
4. calculate the value of x which represents the change in concentration
5. Use the value of x to calculate the final concentration, by plugging in that number to the expressions written for each gas in the E row of the ICE box

SO2(g) + NO2(g) --> NO (g) + SO3(g)
I 0.02 0.04 0.02 0.03
C -x -x +x +x
E 0.02-x 0.04-x 0.02+x 0.03+x

K = [(0.03+x)(0.02+x)]/[(0.04-x)(0.02-x)] = 85.0
x=0.019
SO2=0.02-0.019=0.001 NO2=0.04-0.019=0.021 NO=0.02+0.019=0.039 SO3=0.03+0.019=0.049

Hope this helps!


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