Question 1

Multiple choiceMedium1 mark

For the equilibrium 2A(g)+B(g)⇌2C(g)+D(g)2\text{A}(g)+\text{B}(g)\rightleftharpoons2\text{C}(g)+\text{D}(g) the equilibrium concentrations at a fixed temperature are [A]=0.50[\text{A}]=0.50 [B]=0.20[\text{B}]=0.20 [C]=0.10[\text{C}]=0.10 and [D]=0.10 mol dm−3[\text{D}]=0.10\ \text{mol dm}^{-3} What is KcK_c and which side is favoured?

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Answer: D. The stoichiometric coefficients are the powers in the equilibrium expression:

Kc=[C]2[D][A]2[B]=(0.10)2(0.10)(0.50)2(0.20)=0.020K_c=\frac{[\text{C}]^2[\text{D}]}{[\text{A}]^2[\text{B}]}=\frac{(0.10)^2(0.10)}{(0.50)^2(0.20)}=0.020

(dimensionless). Since Kc<1K_c<1 reactants are favoured at equilibrium.

A: calculates KcK_c correctly but reverses the implication of Kc<1K_c<1
B: omits the square on [C][\text{C}] obtaining 0.200.20 this is still less than 1.
C: uses the reciprocal expression, which gives the equilibrium constant for the reverse reaction.

Remember: a value of Kc<1K_c<1 indicates that reactants predominate at equilibrium.

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