﻿ Answers to Concept Checks - Equilibrium

# EquilibriumAnswers to Concept Checks

6.1

1. First calculate the value of Q from the given concentrations:

Q can now be compared to each value of Keq to predict the direction of the reaction.

 Keq Direction of Reaction ΔG 5.0 × 10−2 At equilibrium: no net reaction 0 5.0 × 10−3 Qc > Keq: proceeds toward reactants (left) Positive 5.0 × 10−1 Qc < Keq: proceeds toward products (right) Negative

2.

3. The concentration of a reactant that converts to product can be considered negligible if it is two or more orders of magnitude less than the initial concentration of the reactant.

 Keq InitialConcentration of A (M) Is the amount reacted negligible? 1 x 10-12 1 Yes 1 x 10-2 0.1 No 1 x 10-3 0.001 No 1 x 10-15 0.001 Yes

4. The first step in solving is to write the equation for Keq for the reaction:

Note that the equation for Keq does not include product D because D is a solid.

Next, using the initial concentrations for A and B and x for the amount that has reacted, plug into the equation for Keq:

Given that Keq = 2.1 x 10—7, the concentrations of A and B are sufficiently large that x can be considered negligible in comparison to both. This allows the equation for Keq to be simplified and solved:

The value of x = 4.2 x 10—9 is equal to both the equilibrium concentration of C and the amount of A and B that have reacted. The approximation that x is negligible compared to the initial concentrations of A and B is valid.

· 6.2

1.

§ Increasing pH of H2SO4 (aq) ⇌ H+ (aq) + HSO4 (aq): [H+] decreases, shifting reaction to the right.

§ Decreasing pressure of 2 C (s) + O2 (g) ⇌ 2 CO (g): Reaction shifts right, favoring the side with more moles of gas.

§ Warming CH4 (g) + 2O2 (g) ⇌ CO2 (g) + 2 H2O (l) + heat: Reaction shifts left, using the additional heat energy to produce more reactants.

§ Removing water from H3PO4 (aq) + H2O (l) ⇌ H3O+ (aq) + H2PO4 (aq): Reaction shifts left. All concentrations would increase proportionately; because there are more products than reactants (and the stoichiometric coefficient is 1 for each reactant and product), the value of Q will increase.

· 6.3

1. Kinetic products are favored at low temperatures with low heat transfer. Thermodynamic products are favored at high temperatures with high heat transfer.

2. Kinetic pathways require a smaller gain in free energy to reach the transition state. They also have a higher free energy of the products, with a smaller difference in free energy between the transition state and the products.

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