Kaplan MCAT General Chemistry Review

The question stem tells us a few things about the reaction: It is combustion, the carbon source has the generic structure C_xH_y, and the products include 10 oxygen atoms, 3 carbon atoms, and 8 H atoms. To balance the reaction, we’d need those atoms on the left side, too. Thus, we find that our unknown sample is actually propane and that we need 5 O₂ molecules.

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.

Takeaways

Consider the number of bonds before applying Hess’s law. Make sure to take note of how many bonds are in a given molecule as well as how many stoichiometric equivalents of that molecule you have.

2) Determine which bonds are broken and which are formed.

C₃H₈: 2 C–C bonds broken, 8 C–H bonds broken

5O₂: 5 O=O bonds broken

3CO₂: 6 C=O bonds formed

4H₂O: 8 O–H bonds formed

Combustion of C₃H₈ will break apart the carbon backbone and the C–H bonds. The carbon is in a straight chain (as opposed to cyclic or branched), so 2 C–C bonds and 8 C–H bonds are broken. For O₂, only one O=O bond is broken. However, we have 5 moles of this reactant, and thus we have 5 O=O bonds broken. Each molecule of carbon dioxide has 2 C=O bonds, but we have 3 moles of CO₂, so we have 6 C=O bonds formed. Similarly, 8 O–H bonds are formed in the 4 moles of water produced.

3) Apply Hess’s law.

H_rxn = H_b (reactants) - H_b (products)

H_rxn = total energy input - total energy released

–2,059 = [2(347) + 8x + 5(497)] - [6(805) + 8(464)]

–2,059 = [3,179 + 8 x] - [8,542]

–2,059 + 8,542 - 3,179 = 8x

x = 413 kJ/mol

Bond dissociation energy is the energy required to break a particular type of bond in one mole of gaseous molecules. Bond energies can be used to estimate the enthalpy of reaction as given by the two equations above. When we start plugging in numbers, we are given all data except for C–H bond enthalpy. We solve for this variable (x in the above equations).

Remember: The equationH_rxn = H_b(reactants) - H_b(products) is simply a restatement of Hess’s law. Bond enthalpy is for bond breaking, and enthalpy of formation, of course, is for bond making. Changing H_Fto H_bswitches the signs and, thus, the order of the equation. Keep in mind that it can also be written as H_rxn = H_b(bonds broken) + H_b(bonds formed), but you must remember to make the bond enthalpies for the products negative because forming bonds releases energy.

4) Use Avogadro’s number.

(413 kJ/mol) × [1 mol/(6.022 × 10²³ molecules)] = 6.86 × 10^-22 kJ/molecule

We see that 413 kJ are found in one mole of C–H bonds. One mole of a substance is equal to 6.022 × 10²³ molecules. Here, we simply use that conversion factor. The result tells us that 6.86 × 10^-22 kJ are stored in each C–H bond.

Things to Watch Out For

There are a number of ways to set up the equation for H_rxn. Whatever equation you use, keep your signs straight. Remember that forming bonds releases energy, whereas breaking bonds requires energy.

Similar Questions

1) Ethanol metabolism in yeast consists of the conversion of ethanol (C₂H₅OH) to acetic acid (CH₃COOH). What is the enthalpy of the reaction if 0.1 mmol of ethanol is metabolized?

2) A second metabolic process involves the net production of 2 ATP and 2 NADH from 2 ADP and 2 NAD⁺. If the conversion of these molecules is endothermic and adds 443.5 kJ to the overall enthalpy of the reaction, find the enthalpy for a “high-energy” phosphate bond.

3) Tristearin is oxidized in the body according to the following reaction: 2 C₅₇H₁₁₀O₆ + 163 O₂ 114 CO₂ + 110 H₂O. If the standard enthalpy for this reaction is -34 MJ mol^-1, find the total enthalpy for the bonds in tristearin.

Heat of Formation

Key Concepts

Chapter 6

Hess’s law:

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

Heat of formation

Combustion

The heat of combustion of glucose (C₆H₁₂O₆) is -2,537.3 kJ/mol. If the H°_f of CO₂ (g) is - 393.5 kJ/mol and the H °_f of H₂O (g) is -241.8 kJ/mol, what is the H°_f of glucose?

1) Write a balanced equation for this reaction.

Unbalanced reaction: C₆H₁₂O₆ + O₂ CO₂ + H₂O