Paper 3 · chemical equilibrium

Nitrogen dioxide decomposes on heating according to the equation: $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$ On introducing 4.8 moles of $NO_2$ to a 1 $dm^3$ container, 0.96 moles of $O_2$ were produced at equilibrium. What is the value of $K_c$ ?

A0.640

B0.480

C0.426

D0.240

Explanation: The marking scheme gives answer C. At equilibrium: moles

O_2

= 0.96, so moles NO produced = 1.92, moles

NO_2

remaining = 4.8 - 1.92 = 2.88. In 1 dm³: [

NO_2

] = 2.88, [NO] = 1.92, [

O_2

] = 0.96.

K_c = \frac{[NO]^2[O_2]}{[NO_2]^2} = \frac{(1.92)^2(0.96)}{(2.88)^2} = \frac{3.6864 \times 0.96}{8.2944} = \frac{3.539}{8.294} \approx 0.427

ZIMSEC Chemistry Paper 3, November 2009, Q5

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