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AP Chemistry : Enthalpy

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Example Questions

Example Question #1 : Enthalpy

Which of the following statements is true concerning a chemical reaction?

Possible Answers:

A catalyst reduces the enthalpy change for the reaction

Endothermic reactions have lower activation energies than exothermic reactions

Exothermic reactions are always spontaneous

The $\small \Delta H$ value for the forward reaction is negative the $\small \Delta H$ value of the reverse reaction

Correct answer:

The $\small \Delta H$ value for the forward reaction is negative the $\small \Delta H$ value of the reverse reaction

Explanation:

When a chemical reaction is represented graphically, we see that the enthalpy change is reversed between the forward and reverse reactions. If a reaction produces energy in a forward process, it will require an input of energy in the reverse process, and vice versa.

$reactant\ +heat\rightleftharpoons product,\ \Delta H=+$

$product\rightleftharpoons heat\ +reactant,\ \Delta H=-$

A catalyst only affects the rate of a chemical reaction; it does not affect the equilibrium. Finally, exothermic reactions are not always spontaneous, but will have lower activation of energies compared to endothermic reactions.

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Example Question #1 : Enthalpy

$\Delta H_f=0kJ$

O_2 $\Delta H_f=0kJ$

MgO $\Delta H_f=-601kJ$

What is the change in enthalpy for the following reaction?

$2Mg +O_2\rightarrow 2MgO$

Possible Answers:

$601\frac{kJ}{mol}$

$-601\frac{kJ}{mol}$

$-300.5\frac{kJ}{mol}$

$1202\frac{kJ}{mol}$

$-1202\frac{kJ}{mol}$

Correct answer:

$-1202\frac{kJ}{mol}$

Explanation:

The change in enthalpy is calculated by:

$\Delta H_{products} - \Delta H_{reactants}$

When $\Delta H$ cannot be measured, it can be calculated from known enthalpies of formation.

$\Delta H_f=0kJ$

O_2 $\Delta H_f=0kJ$

MgO $\Delta H_f=-601kJ$

It is important to first balance the reaction before performing calculations. The coefficients are important in determining the change in enthalpy of a reaction.

$2Mg +O_2\rightarrow 2MgO$

$\Delta H_{rxn}=2(\Delta H_{MgO})-2(\Delta H_{Mg})-\Delta H_{O_{2}}$

$\Delta H_{rxn}=2(-601kJ)-2(0kJ)-0kJ$

$\Delta H_{rxn}=-1202\frac{kJ}{mol}$

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Example Question #1 : Enthalpy

The formation of nitrogen dioxide is a two step process.

$\small N_{2} + O_{2} \rightarrow 2NO$ $\small \Delta H = 180kJ$

$\small 2NO + O_{2}\rightarrow 2NO_{2}$ $\small \Delta H = -112kJ$

The net reaction is $\small N_{2} + 2O_{2} \rightarrow 2NO_{2}$ .

What is the change in enthalpy when creating four moles of nitrogen dioxide?

Possible Answers:

$\small 68kJ$

$\small 292kJ$

$\small -44kJ$

$\small 136kJ$

173kJ

Correct answer:

$\small 136kJ$

Explanation:

Hess's law states that the change in enthalpy for a total reaction can be considered equal to the sum of the enthalpy changes for every step involved in the reaction. In other words, we can determine the enthalpy change for nitrogen dioxide by adding the enthalpy changes for both steps involved in its formation.

$\Delta H_{step1}+\Delta H_{step2}=\Delta H_{}$

180kJ+(-112kJ)=68kJ

This gives us the total change in enthalpy for the listed reaction, $\small N_{2} + 2O_{2} \rightarrow 2NO_{2}$ . Because the question asks for the enthalpy change for four moles of nitrogen dioxide, the value must be doubled. The reaction only produces two moles of nitrogen dioxide.

$\frac{68kJ}{2mol\ NO_2}*2=\frac{136kJ}{4mol\ NO_2}$

$\Delta H_{4mol}=136kJ$

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Example Question #1 : Enthalpy

Calculate ΔH for the following reaction:

CH_{4 (g)} + O₂ (g) ⇌ CO_{2 (g)} + H₂O (l)

Compound ΔH

CH₄ (g) -74.8 kJ/mol

H₂O (l) -285.8 kJ/mol

CO₂(g) -393.5 kJ/mol

Possible Answers:

604.5 kJ/mol

-604.5 kJ/mol

-890.3 kJ/mol

889.7 kJ/mol

not enough information

Correct answer:

-890.3 kJ/mol

Explanation:

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔHf O₂ or any element is 0

First step is to balance the equation:

CH_{4 (g)} + O₂ (g) ⇌ CO_{2 (g)} + 2H₂O (l)

ΔH = Σ ΔHf products - Σ ΔHf reactants

= [-393.5 kJ/mol + 2(-285.8) kJ/mol] - (-74.8 kJ/mol)

= -890.3 kJ/mol

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Example Question #2 : Enthalpy

$A \rightarrow E+F$ $\Delta H=-50kJ$

$E+F\rightarrow C+G$ $\Delta H=+85 kJ$

$B+G \rightarrow D$ $\Delta H= -10kJ$

What is the change in enthalpy for the given reaction?

$A+B\rightarrow C+D$

Possible Answers:

-30kJ

+45kJ

+75kJ

-100kJ

+25kJ

Correct answer:

+25kJ

Explanation:

The change in enthalpy is calculated by:

$\Delta H_{products} - \Delta H_{reactants}$

When $\Delta H$ cannot be measured, it can be calculated from known reactions. In this case the known reactions are given.

$A \rightarrow E+F$ $\Delta H=-50kJ$

$E+F\rightarrow C+G$ $\Delta H=+85 kJ$

$B+G \rightarrow D$ $\Delta H= -10kJ$

Since the reactions are in the correct order, adding all the $\Delta H$ values together can be used to calculate the $\Delta H$ of the reaction.

$A+E+F+B+G\rightarrow E+F+C+G+D$

$\text{Net reaction: }A+B\rightarrow C+D$

$\Delta H=-50kJ+85kJ-10kJ=25kJ$

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Example Question #6 : Enthalpy

NO_2 $\Delta H_f=33kJ$

$O_{2(g)}$ $\Delta H_f=0kJ$

$N_{2(g)}$ $\Delta H_f=0kJ$

What is the change in enthalpy for the following reaction?

$N_{2(g)}+O_{2(g)}\rightarrow NO_2$

Possible Answers:

$33\frac{kJ}{mol}$

$66\frac{kJ}{mol}$

$15\frac{kJ}{mol}$

$25\frac{kJ}{mol}$

$42\frac{kJ}{mol}$

Correct answer:

$66\frac{kJ}{mol}$

Explanation:

The change in enthalpy is calculated by:

$\Delta H_{products} - \Delta H_{reactants}$

When $\Delta H$ cannot be measured, it can be calculated from known enthalpies of formation.

NO_2 $\Delta H_f=33kJ$

$O_{2(g)}$ $\Delta H_f=0kJ$

$N_{2(g)}$ $\Delta H_f=0kJ$

It is important to first balance the reaction before performing calculations. The coefficients are important in determining the change in enthalpy of a reaction.

$N_{2(g)}+2O_{2(g)}\rightarrow 2NO_2$

$\Delta H_{rxn}=2(\Delta H_{NO_2})-\Delta H_{N_{(g)}}-2(\Delta H_{O_{2(g)}})$

$\Delta H_{rxn}=2(33kJ)-0kJ-2(0kJ)$

$\Delta H_{rxn}=66\frac{kJ}{mol}$

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Example Question #7 : Enthalpy

$N_{2(g)}$ $\Delta H_f=0 kJ$

$O_{2(g)}$ $\Delta H_f=0 kJ$

$\Delta H_f=90kJ$

What is the enthalpy of the following reaction?

$N_{2(g)}+O_{2(g)}\rightarrow 2NO$

Possible Answers:

$180\frac{kJ}{mol}$

$15\frac{kJ}{mol}$

$90\frac{kJ}{mol}$

$45\frac{kJ}{mol}$

$0\frac{kJ}{mol}$

Correct answer:

$180\frac{kJ}{mol}$

Explanation:

The change in enthalpy is calculated by:

$\Delta H_{products} - \Delta H_{reactants}$

When $\Delta H$ cannot be measured, it can be calculated from known enthalpies of formation.

$N_{2(g)}$ $\Delta H_f=0 kJ$

$O_{2(g)}$ $\Delta H_f=0 kJ$

$\Delta H_f=90kJ$

It is important to first balance the reaction before performing calculations. The coefficients are important in determining the change in enthalpy of a reaction. For this particular reaction, since there are two moles of product, the enthalpy of formation for must be multiplied by two.

$N_{2(g)}+O_{2(g)}\rightarrow 2NO$

$\Delta H_{rxn}=2(\Delta H_{NO})-\Delta H_{N_{2(g)}}-\Delta H_{O_{2(g)}}$

$\Delta H_{rxn}=2(90kJ)-0kJ-0kJ$

$\Delta H_{rxn}=180\frac{kJ}{mol}$

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Example Question #8 : Enthalpy

$NO_{(g)}$ $\Delta H_f=90kJ$

$O_{2(g)}$ $\Delta H_f=0kJ$

$NO_{2}$ $\Delta H_f=33kJ$

What is the change in enthalpy for the following reaction?

$NO_{(g)}+O_{2(g)} \rightarrow NO_2$

Possible Answers:

$-147\frac{kJ}{mol}$

$-114\frac{kJ}{mol}$

$123\frac{kJ}{mol}$

$-24\frac{kJ}{mol}$

$-57\frac{kJ}{mol}$

Correct answer:

$-114\frac{kJ}{mol}$

Explanation:

The change in enthalpy is calculated by:

$\Delta H_{products} - \Delta H_{reactants}$

When $\Delta H$ cannot be measured, it can be calculated from known enthalpies of formation.

$NO_{(g)}$ $\Delta H_f=90kJ$

$O_{2(g)}$ $\Delta H_f=0kJ$

$NO_{2}$ $\Delta H_f=33kJ$

It is important to first balance the reaction before performing calculations. The coefficients are important in determining the change in enthalpy of a reaction.

$2NO_{(g)}+O_{2(g)}\rightarrow 2NO_{2}$

$\Delta H_{rxn}=2(\Delta H_{NO_2})-2(\Delta H_{NO_{(g)}})-\Delta H_{O_{2(g)}}$

$\Delta H_{rxn}=2(33kJ)-2(90kJ)-0kJ$

$\Delta H_{rxn}=-114\frac{kJ}{mol}$

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Example Question #9 : Enthalpy

$C_4H_{10}$ $\Delta H_f=-126kJ$

O_2 $\Delta H_f=0kJ$

CO_2 $\Delta H_f=-394kJ$

H_2O $\Delta H_f=-286kJ$

What is the change in enthalpy for the following reaction?

$C_4H_{10}+O_2\rightarrow CO_2+H_2O$

Possible Answers:

$-145\frac{kJ}{mol}$

$-554\frac{kJ}{mol}$

$-5760\frac{kJ}{mol}$

$-500\frac{kJ}{mol}$

$+1416\frac{kJ}{mol}$

Correct answer:

$-5760\frac{kJ}{mol}$

Explanation:

The change in enthalpy is calculated by:

$\Delta H_{products} - \Delta H_{reactants}$

When $\Delta H$ cannot be measured, it can be calculated from known enthalpies of formation.

$C_4H_{10}$ $\Delta H_f=-126kJ$

O_2 $\Delta H_f=0kJ$

CO_2 $\Delta H_f=-394kJ$

H_2O $\Delta H_f=-286kJ$

It is important to first balance the reaction before performing calculations. The coefficients are important in determining the change in enthalpy of a reaction.

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

$\Delta H_{rxn}=8(\Delta H_{CO_2})+10(\Delta H_{H_2O})-2(\Delta H_{C_4H_{10}})-13(\Delta H_{O_2})$

$\Delta H_{rxn}=8(-394kJ)+10(-286kJ)-2(-126kJ)-13(0kJ)$

$\Delta H_{rxn}=-5760\frac{kJ}{mol}$

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Example Question #10 : Enthalpy

Consider the following combustion reaction:

$C_{3}H_{8}_{(g)} + 5O_{2}_{(g)} \rightarrow 3CO_{2}_{(g)} + 4H_{2}O_{(g)}$

The following list is the enthalpies of formation for the compounds:

$\Delta H_{C_{3}H_{8}} = -103.85\frac{kJ}{mol}$

$\Delta H_{CO_{2}} = -393.5\frac{kJ}{mol}$

$\Delta H_{H_{2}O} = -241.82\frac{kJ}{mol}$

If one mole of propane is burned, what is the enthalpy of the reaction?

Possible Answers:

-2044 kJ

-739kJ

-531kJ

-2148kJ

Correct answer:

-2044 kJ

Explanation:

Since we have the enthalpies of formation, we can find the enthalpy of the reaction using the following equation:

$\Delta H_{reaction} = \Delta H_{products} - \Delta H_{reactants}$

We will need to use the coefficients from the balanced equation to calculate the enthalpy.

$C_{3}H_{8}_{(g)} + 5O_{2}_{(g)} \rightarrow 3CO_{2}_{(g)} + 4H_{2}O_{(g)}$

Since oxygen gas is elemental, it does not have an enthalpy of formation, and is omitted from the equation.

$\Delta H_{reaction}=3(\Delta H_{CO_2})+4(\Delta H_{H_2O})-\Delta H_{C_3H_8}$

$\Delta H_{reaction}=3(-393.5\frac{kJ}{mol})+4(-241.82\frac{kJ}{mol})-(-103.85\frac{kJ}{mol})$

$\Delta H_{reaction}=-2044\frac{kJ}{mol}$

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AP Chemistry : Enthalpy

Study concepts, example questions & explanations for AP Chemistry

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