MCAT General Chemistry Review  Alexander Stone Macnow, MD 20192020
Equilibrium
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.
K_{eq} 
Direction of Reaction 
ΔG 
5.0 × 10^{−2} 
At equilibrium: no net reaction 
0 
5.0 × 10^{−3} 
Q_{c} > K_{eq}: proceeds toward reactants (left) 
Positive 
5.0 × 10^{−1} 
Q_{c} < K_{eq}: 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.

Initial 
Is the amount 
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 H_{2}SO_{4} (aq) ⇌ H^{+} (aq) + HSO_{4}^{−} (aq): [H^{+}] decreases, shifting reaction to the right.
§ Decreasing pressure of 2 C (s) + O_{2} (g) ⇌ 2 CO (g): Reaction shifts right, favoring the side with more moles of gas.
§ Warming CH_{4} (g) + 2O_{2} (g) ⇌ CO_{2} (g) + 2 H_{2}O (l) + heat: Reaction shifts left, using the additional heat energy to produce more reactants.
§ Removing water from H_{3}PO_{4} (aq) + H_{2}O (l) ⇌ H_{3}O^{+} (aq) + H_{2}PO_{4}^{−} (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.