Kaplan MCAT General Chemistry Review

Balanced reaction: C₆H₁₂O₆ + 6 O₂ 6 CO₂ + 6 H₂O

The unbalanced reaction above is typical of all hydrocarbon combustion reactions. (Unless otherwise noted, presume that combustion of carbohydrates is with oxygen gas.) Begin by balancing the carbons on the left side (6 CO₂), then balance the hydrogens on the left side (12 H₂O), and conclude by balancing the oxygen gas on the right side (6 O₂).

Remember: For our purposes, it is completely acceptable to have a fractional coefficient in front of a diatomic molecule: 2 C₂H₄ + (7/2) O₂3 H₂O + 2 CO₂is equivalent to 4 C₂H₄ + 7 CO₂6 H₂O + 4 CO₂, but the math is simpler for the former.

2) Apply Hess’s law.

H_rxn = H_F (products) - H_F (reactants)

-2,537.3 = [6(-393.5) + 6(-241.8)] - [H_F (glucose)]

Rearranging to solve for H_F (glucose):

H_f (glucose) = 2,537.3 + [6(-393.5) + 6(-241.8)]

H_f (glucose) = 2,537.3 + [-2,361 + -1,450.8]

H_f (glucose) = 2,537.3 + [-3,811.8]

H_f (glucose) = -1,274.5

The heat of formation is defined as the heat absorbed or released during the formation of a pure substance from the elements at a constant pressure. Therefore, by definition, diatomic gases like oxygen have a heat of formation of zero. A negative heat of formation means that heat is released to form the product, whereas a positive heat of formation means that heat is required to form the product. The overall combustion reaction of glucose releases 2,537.3 kJ/mol of heat.

Takeaways

Always identify the balanced equation for the reaction before you begin to apply Hess’s law. There is a second way of thinking about Hess’s law that may be applicable in some questions as well (see previous topic). The given information in the passage and/or question stem will dictate which equation to use. Finally, recall that enthalpy is a state function, and regardless of the path you take to get from the reactants to the products, the change in enthalpy will be the same.

Things to Watch Out For

At least one of the wrong answer choices for thermochemistry questions will be a result of carelessness with signs. Organized scratchwork in a stepwise fashion will facilitate avoiding this problem, but perhaps more important is maintaining the ability to approximate the answer. Only experience (aka practice!) will breed such wisdom.

Similar Questions

1) Given the H_F of carbon dioxide and water, what other piece(s) of information must you have to calculate the H_comb of ethane?

2) If the H_F of acetylene is 226.6 kJ/mol, what is the H_comb of acetylene?

3) If the H_F of NaBr (s) is -359.9 kJ/mol, what is the sum of each H_F of the following series of five reactions?

Na (s) Na (g) Na⁺ (g)

Br₂(g) Br(g) Br^- (g)

Na⁺(g)+Br^-(g)NaBr(s)

Partial Pressures

Key Concepts

Chapter 7

Dalton’s law

P_A = X_AP_Total (atm)

Mole fraction

32 g of oxygen, 28 g of nitrogen, and 22 g of carbon dioxide are confined in a container with partial pressures of 2 atm, 2 atm, and 1 atm respectively. A student added 57 g of a halogen gas to this container and observed that the total pressure increased by 3 atm. Can you identify this gas?

1) Determine the number of moles for each gas.

This is a more complicated style of partial pressure questions, yet the first step is still the basic one of identifying the number of moles for each gas.

2) Solve for the relevant variable.

All partial pressure questions boil down to this formula. The relevant variable here is the mole fraction of the halogen gas. The partial pressure of the gas is 3 atm, and the total pressure is 8 atm.

Takeaways

Partial pressure questions will require manipulation of the formulas above, so the key is always to keep track of what is given to you and what the question is asking for.

3) Use the mole fractionX_Ato solve for the number of moles and the MW of the halogen gas.

The mole fraction of a substance is the number of moles of the substance as a fraction of the total number of moles in the container:

Rearranging the formula, we have # moles of A = (X_A)(total # of moles). Note that the total number of moles is not known, but we can express it algebraically as 2.5 + n_A, where n_A is defined as the number of moles of A.

# moles of A = (X_A)(total # of moles)

Things to Watch Out For

Dalton’s law assumes that the gases do not react with each other.

This MW corresponds to F₂. Of course, on Test Day you will roughly round such that 60 g = 1.5 mol and look for the halogen gas using your calculated MW of 40 g/mol. Again, the only gas possible is F₂.

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