Kaplan MCAT General Chemistry Review

2) If a catalyst were added to the reaction of A going to C, as in step 3 above, would the energies of A and C be changed as a result? Why or why not?

3) There are actually intermediates involved in the reactions producing both B and C. These intermediates are shown below. Sketch how each reaction profile would look, including the involvement of these intermediates. Be sure to indicate which intermediate is relatively more stable.

Thermodynamic Equilibrium

Key Concepts

Chapter 6

Thermochemistry

Gibbs free energy

Enthalpy

Entropy

Equilibrium constant, K_eq

Reaction quotient, Q

G = H - (kJ/mol)

G° = -RT ln K_eq (kJ/mol)

G = G° + RT ln Q (kJ/mol)

The reaction 2 NO ( g) + Cl₂ ( g) 2 NOCl ( g) adheres to the following thermodynamic data:

H -77.1 kJ/molS -121 J/KG -44.0 kJ/molK_eq1.54 × 10⁷

Suppose that, in equilibrium, NO exerts 0.6 atm of pressure and Cl₂ adds 0.3 atm; find the partial pressure of NOCl in this equilibrium. Also, find the temperature at which the thermodynamic data in the table were reported. K_eq is related to K_p by the following equation: K_p = K_eq(RT)ⁿ where n is the change in number of moles of gas evolved as the reaction moves forward. (R = 8.314 J/K · mol.)

1) Find the temperature at which the thermodynamic data are true.

G = H -

(-44 kJ/mol) = (-77 kJ/mol) - T (-0.121 kJ/K × mol) T = 273 K

Being able to work with the equation G = H - is absolutely crucial for Test Day. Specifically concerning the data here, because both H and S are negative, the reaction will become “less spontaneous” as we increase temperature. This will help narrow down our answer choices on Test Day.

MCAT Pitfall: Notice that not all of the state functions were given in the same unit! Had you blindly put in entropy without changing its units, you would have obtained a temperature near absolute zero (0 K). At absolute zero, molecules no longer move, and it is unlikely that this reaction would have such a high equilibrium constant.

Takeaways

Two equations should get you through nearly any thermochemistry question. Remember to round your numbers and to predict the ballpark for your answers wherever possible.

2) Findn.

For the equation

2 NO ( g) + Cl₂ ( g) 2 NOCl ( g),

n = 2 - (2 + 1) = -1.

Things to Watch Out For

Pay close attention to the units used.

3) Find K_p.

K_p = K_eq(RT )ⁿ

K_p = (1.54 × 10⁷)(8.314 J/K·mole)(273 K)^-1 = 6,785

Use the temperature value from step #1.

You are not responsible for memorizing the equation. However, in the MCAT, you have to be able to use a brand-new equation to solve for the answer.

4) Find the partial pressure.

Plug in the given data into the reaction quotient.

Similar Questions

1) If K_eq = 7.4 × 10^-3 for CH₄ (g)+ 2 H₂O (g)? CO₂ (g)+ 4 H₂ (g),which is more plentiful, the reactants or the products?

2) If pyrophosphoric acid (H₄P₂O₇) and arsenous acid (H₃AsO₃) have acid dissociation constants of 3 × 10^-2 and 6.6 × 10^-10, respectively, at room temperature, find the Gibbs free energy of each dissociation reaction and determine if it is spontaneous. What does this mean for the H and S for these reactions?

3) A chemist is given three liquid-filled flasks, each labeled with generic thermodynamic data. She is told to put one in a cold room, to put one on a Bunsen burner, and to leave one on the benchtop—whatever conditions will best facilitate the reaction. If the flasks are labeled as follows, which flask goes where?

A H < 0, S > 0

E H < 0, S < 0

P H > 0, S > 0

Bond Enthalpy

Key Concepts

Chapter 6

Hess’s law:

H_rxn = H_F(products) - H_F(reactants)(kJ/mol)

Enthalpy

Bond dissociation energy

Combustion

Stoichiometry

H_rxn = H_b (reactants) - H_b (products) (kJ/mol)

H_rxn = total energy input - total energy released (kJ/mol)

An unknown compound containing only carbon and hydrogen is subjected to a combustion reaction in which 2,059 kJ of heat are released. If 3 moles of CO₂ and 4 moles of steam are produced for every mole of the unknown compound reacted, find the enthalpy for a single C–H bond.

BondBond Dissociation Energy (kJ/mol)O=O497C=O805O–H464O–H347

1) Write a balanced equation for this reaction.

C₃H₈ + 5O₂ 3CO₂ + 4H₂O