Question 17.14: Determining Equilibrium Parameters from Molecular Scenes Pro......
Determining Equilibrium Parameters from Molecular Scenes
Problem For the reaction,
X(g) + Y_2(g) \xrightleftharpoons[] XY(g) + Y(g) ΔH > 0the following molecular scenes depict different reaction mixtures (X is green, Y is purple) (Fig 17.14):
(a) If K = 2 at the temperature of the reaction, which scene represents the mixture at equilibrium? (b) Will the reaction mixtures in the other two scenes proceed toward reactants or toward products to reach equilibrium? (c) For the mixture at equilibrium, how will a rise in temperature affect [Y_2]?
Plan (a) We are given the balanced equation and K and must choose the scene that represents the mixture at equilibrium. We write Q, and for each scene, count particles and find the value of Q. Whichever scene gives a Q equal to K (that is, equal to 2) represents the mixture at equilibrium. (b) For each of the other two reaction mixtures, we compare the value of Q with 2. If Q > K, the numerator (product side) is too high, so the reaction proceeds toward reactants; if Q < K, the reaction proceeds toward products. (c) We are given that ΔH > 0, so we must see whether a rise in T increases or decreases [Y_2], one of the reactants.
Solution (a) For the reaction, we have Q = \frac{[XY][Y]}{[X][Y_2]}. Thus,
\text{scene 1 : }Q = \frac{5 × 3}{1 × 1} = 15 \text{ scene 2 : }Q = \frac{4 × 2}{2 × 2} = 2 \text{scene 3 : }Q = \frac{3 × 1}{3 × 3} = \frac{1}{3}For scene 2, Q = K, so scene 2 represents the mixture at equilibrium.
(b) For scene 1, Q (15) > K (2), so the reaction proceeds toward reactants.
For scene 3, Q (\frac{1}{3}) < K (2), so the reaction proceeds toward products.
(c) The reaction is endothermic, so heat acts as a reactant:
X(g) + Y_2(g) + heat \xrightleftharpoons[] XY(g) + Y(g)
Therefore, adding heat shifts the reaction to the right, so [Y_2] decreases.
Check (a) Remember that quantities in the numerator (or denominator) of Q are multiplied, not added. For example, the denominator for scene 1 is 1 × 1 = 1, not 1 + 1 = 2.
(c) A good check is to imagine that ΔH < 0 and see if you get the opposite result:
X(g) + Y_2(g) \xrightleftharpoons[] XY(g) + Y(g) + heat
If ΔH < 0, adding heat would shift the reaction to the left and increase [Y_2].