Answers 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 K_eq 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	Initial Concentration 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 K_eq for the reaction:

   Note that the equation for K_eq 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 K_eq:  

   Given that K_eq = 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 K_eq 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.