## Chapter 17

## Q. 17.14

Determining Equilibrium Parameters from Molecular Scenes

**Problem** For the reaction,

the 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]?

## Step-by-Step

## Verified Solution

**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,

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].