At 2000 °C the equilibrium constant for the reaction below is Kc= 2.4x103 . If the initial concentration of NO is 0.500 M, what are the equilibrium concentrations of each substance?
2 NO (g) ⇌ N2 (g) + O2 (g)
Solution:-
NO (g) | N2 (g) | O2 (g) | |
I | 0.500 | 0 | 0 |
C | -2x | +x | +x |
E | (0.500 - 2x) | x | x |
Now,
Kc = [N2] [O2] / [NO]2
(2.4 × 103) = (x.x) / (0.500 - 2x)2
(2.4 × 103) = x2 / (0.500 - 2x)2
On taking square roots on both sides, we get
48.99 = x / (0.500 - 2x)
24.49 - 97.98 x = x
24.49 = 98.98 x
x = (24.49) / (98.98) = 0.2474 M
x = [N2] = [O2] = 0.2474 M
[NO] = (0.500 - 2x) = [0.500 - 2 (0.2474)] = 0.0052 M
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