Enthalpy - AP Chemistry

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Question

$2\text{K} + \text{Cl}_2 \rightarrow 2\text{KCl} + 640\text{ kJ}$

Suppose that 1.5 moles of potassium reacts completely with an excess of chlorine gas. What is the enthalpy change of the reaction?

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Answer

First, note that heat is a product of the reaction. This means that the reaction releases heat and is exothermic, so the enthalpy change must be negative. Next, consider that 2 moles of sodium would react to produce 640 kJ. Thus, 1.5 moles of sodium would react to produce

$640\text{ kJ}\times\frac{1.5\text{ mol}}{2\text{ mol}}=480\text{ kJ}$ .

Hence, the enthalpy change is $-480\text{ kJ}$ .

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Question

H2O ice melts to liquid H2O \[Heat is added\]

Which of the following is correct?

I. Δ H is positive

II. ΔS is positive

III. ΔH is negative

IV. ΔS is negative

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Answer

When heat is added, enthalpy increases, ΔH refers to enthalpy

ΔS refers to entropy, which is disorder; disorder increases as it moves from solid to liquid to gas state.

As ice goes to liquid water, entropy increases

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Question

Calculate ΔH for the following reaction:

CH4 (g) + O2 (g) ⇌ CO2 (g) + H2O (l)

Compound ΔH

CH4 (g) -74.8 kJ/mol

H2O (l) -285.8 kJ/mol

CO2 (g) -393.5 kJ/mol

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Answer

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔHf O2 or any element is 0

First step is to balance the equation:

CH4 (g) + O2 (g) ⇌ CO2 (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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Question

Which of the following is a true statement regarding the energy involved in the formation of Methane, CH4, from graphite C(s) and H2(g) ?

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Answer

This problem requires a knowledge of the energy involved when bonds are broken versus when bonds are formed. Energy is released when bonds are formed, and energy is consumed when bonds are broken. Here, methane is being formed from C and H2. Since the carbon source, graphite, is monoatomic the carbon will be involved only in the formation of C–H bonds on it's way to forming methane. The H however exists as H2 so bonds will need to be broken to allow for C–H bonds to form. So first energy is consumed to break the H–H bonds, it is then released when the C–H bonds form.

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Question

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?

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Answer

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.

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

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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Question

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

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Answer

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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Question

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

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

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

What is the change in enthalpy for the given reaction?

$A+B\rightarrow C+D$

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Answer

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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Question

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

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

$\Delta H_f=90kJ$

What is the enthalpy of the following reaction?

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

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Answer

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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Question

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

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

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

What is the change in enthalpy for the following reaction?

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

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Answer

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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Question

$\Delta H_f=33kJ$

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

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

What is the change in enthalpy for the following reaction?

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

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Answer

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.

$\Delta H_f=33kJ$

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

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

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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Question

$\Delta H_f=0kJ$

$\Delta H_f=0kJ$

$\Delta H_f=-601kJ$

What is the change in enthalpy for the following reaction?

$2Mg +O_2\rightarrow 2MgO$

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Answer

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.

$\Delta H_f=0kJ$

$\Delta H_f=0kJ$

$\Delta H_f=-601kJ$

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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Question

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

$\Delta H_f=0kJ$

$\Delta H_f=-394kJ$

$\Delta H_f=-286kJ$

What is the change in enthalpy for the following reaction?

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

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Answer

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.

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

$\Delta H_f=0kJ$

$\Delta H_f=-394kJ$

$\Delta H_f=-286kJ$

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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Question

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?

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Answer

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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Question

Which of the following equations is correct for finding change in enthalpy?

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Answer

The change in enthalpy of a reaction is determined by the amount of heat gained when the products bonds form, versus the amount of heat lost to break the bonds in the reactants. The correct formula for the enthalpy of reaction is the difference between the enthalpy of the products and that of the reactants:

$\Delta H _{products} - \Delta H _{reactants}$

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Question

$S_{(s)} + O_{2(g)} \rightarrow SO_{2(g)}$ $\Delta H = x$

$S_{(s)} + \frac{3}{2}O_{2(g)} \rightarrow SO_{3(g)}$ $\Delta H = y$

Based on the given reactions, what is the enthalpy change in the following reaction?

$2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)}$

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Answer

The reactions given show the heats of formation of all the reactants and products, and we know the heat of formation of $O_{2}$ is zero because it is a pure element. Formation of the product, $SO_{3}$ gives us . For the reactants, $SO_{2}$ gives us plus for $O_{2}$ . Recall that the overall reaction's change in enthalpy is:

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

Thus for this overall reaction:

$\Delta H_{reaction} = 2y-2x$

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Question

H2O ice melts to liquid H2O \[Heat is added\]

Which of the following is correct?

I. Δ H is positive

II. ΔS is positive

III. ΔH is negative

IV. ΔS is negative

Tap to reveal answer

Answer

When heat is added, enthalpy increases, ΔH refers to enthalpy

ΔS refers to entropy, which is disorder; disorder increases as it moves from solid to liquid to gas state.

As ice goes to liquid water, entropy increases

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Question

Calculate ΔH for the following reaction:

CH4 (g) + O2 (g) ⇌ CO2 (g) + H2O (l)

Compound ΔH

CH4 (g) -74.8 kJ/mol

H2O (l) -285.8 kJ/mol

CO2 (g) -393.5 kJ/mol

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Answer

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔHf O2 or any element is 0

First step is to balance the equation:

CH4 (g) + O2 (g) ⇌ CO2 (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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Question

Which of the following is a true statement regarding the energy involved in the formation of Methane, CH4, from graphite C(s) and H2(g) ?

Tap to reveal answer

Answer

This problem requires a knowledge of the energy involved when bonds are broken versus when bonds are formed. Energy is released when bonds are formed, and energy is consumed when bonds are broken. Here, methane is being formed from C and H2. Since the carbon source, graphite, is monoatomic the carbon will be involved only in the formation of C–H bonds on it's way to forming methane. The H however exists as H2 so bonds will need to be broken to allow for C–H bonds to form. So first energy is consumed to break the H–H bonds, it is then released when the C–H bonds form.

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Question

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?

Tap to reveal answer

Answer

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.

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

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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Question

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

Tap to reveal answer

Answer

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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