Chemical Equilibrium Post Assessment 3

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Jacinda Wollenweber 1D
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Chemical Equilibrium Post Assessment 3

Postby Jacinda Wollenweber 1D » Sun Jul 23, 2017 8:34 pm

16. 0.482 mol N2 and 0.933 mol O2 are placed in a 10.0 L reaction vessel and form N2O (dinitrogen oxide): 2N2(g) + O2(g) ⇌ 2N2O(g) KC = 2.0 x 10-37
What is the composition of the equilibrium mixture?

A. [N2] = 0.0482 mol.L-1; [O2] = 0.0933 mol.L-1; [N2O] = 6.6 x 10-21 mol.L-1

B. [N2] = 0.0933 mol.L-1; [O2] = 0.0482 mol.L-1; [N2O] = 6.6 x 10-21 mol.L-1

C. [N2] = 0.482 mol.L-1; [O2] = 0.933 mol.L-1; [N2O] = 6.6 x 10-21 mol.L-1

D. [N2] = 6.6 x 10-21 mol.L-1; [O2] = 0.0933 mol.L-1; [N2O] = 0.0482 mol.L-1

I am not getting the right answer for N20, so can someone guide me through the steps please. Thanks!

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Re: Chemical Equilibrium Post Assessment 3

Postby Chem_Mod » Mon Jul 24, 2017 9:46 am

INITIAL MOLAR CONCENTRATION OF N2 = 0.482 mol/10.0 L = 0.0482 mol.L-1
INITIAL MOLAR CONCENTRATION OF O2 = 0.933 mol/10.0 L = 0.0933 mol.L-1

ASSUME THAT O2 CHANGES ITS MOLAR CONCENTRATION BY -X
N2 O2 N2O
INITIAL MOLAR CONCENTRATION 0.0482 0.0933 0
CHANGE IN MOLAR CONCENTRATION -2X -X +2X
EQUILIBRIUM MOLAR CONCENTRATION 0.0482 - 2X 0.0933 - X 2X

K = [N2O]2/[N2]2 [O2] = (2X)2/(0.0482 - 2X)2 (0.0933 - X)
(which is a cubic equ with x3, and difficult to solve)

HOWEVER, SINCE K = 2.0 x 10-37 WHICH IS V.SMALL WE CAN APPROXIMATE
0.0482 - 2X = 0.0482 AND 0.0933 - X = 0.0933

 K = 4X2/0.04822 x (0.0933) = 2.0 x 10-37

X = 3.3 x 10-21
(APPROXIMATION IS VALID AS X IS VERY MUCH SMALLER THAN THE INITIAL VALUES)

 THE COMPOSITION OF THE EQUILIBRIUM MIXTURE IS:
[N2] = 0.0482 - 2X = 0.0482 mol.L-1
[O2] = 0.0933 - X = 0.0933 mol.L-1
[N2O] = 2X = 6.6 x 10-21mol.L-1


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