5. Given the following reaction at equilibrium, if K c = 6.14 x 10 5 at 230.0 °C

Question

5. Given the following reaction at equilibrium, if Kc = 6.14 x 105 at 230.0 °C, Kp =------

   2NO (g) + O2 (g) INCLUDEPICTURE "http://session.masteringchemistry.com/problemAsset/1801237/2/doublearrowshort.jpg" * MERGEFORMATINET 2NO2(g)

   [A] 1.49 104 [B] 3.67 10-2 [C] 6.44 105 [D] 2.54 107 [E] 2.53 106

   6. The value of Keq for the equilibrium

H2 (g) + I2 (g) INCLUDEPICTURE "http://session.masteringchemistry.com/problemAsset/1801224/1/doublearrowshort.jpg" * MERGEFORMATINET 2HI (g)

is 794 at 25 °C. At this temperature, what is the value of Keq

for the equilibrium below?

HI (g) INCLUDEPICTURE "http://session.masteringchemistry.com/problemAsset/1801224/1/doublearrowshort.jpg" * MERGEFORMATINET 1/2 H2 (g) + 1/2 I2 (g)

[A] 397 [B] 1588 [C] 28 [D] 0.0013 [E] 0.035

Explanation / Answer

1) 2NO(g) + O2(g) <-------> 2NO2(g)

Kp = Kc*(R*T)n ; where n = moles of gaseous products - moles of gaseous reactants = -1

Thus, Kp = 6.14*105/(8.314*503) = 146.82

2) H2(g) + I2(g) ----> 2HI(g) ; Kc = 794

Kc = [HI]2/{[H2]*[I2]} = 794..........(1)

Now, for the REACTION:-

(1/2)H2(g) + (1/2)I2(g) ----> HI(g)

Kc = [HI]/{[H2]*[I2]}1/2 = 7941/2 = 28.178

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