Chapter 10: Thermochemistry

Learning Outcomes

  • Explain Hess’s law and use it to compute reaction enthalpies

Hess’s Law

There are two ways to determine the amount of heat involved in a chemical change: measure it experimentally, or calculate it from other experimentally determined enthalpy changes. Some reactions are difficult, if not impossible, to investigate and make accurate measurements for experimentally. And even when a reaction is not hard to perform or measure, it is convenient to be able to determine the heat involved in a reaction without having to perform an experiment.

This type of calculation usually involves the use of Hess’s law, which states: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. Hess’s law is valid because enthalpy is a state function: Enthalpy changes depend only on where a chemical process starts and ends, but not on the path it takes from start to finish. For example, we can think of the reaction of carbon with oxygen to form carbon dioxide as occurring either directly or by a two-step process. The direct process is written:

[latex]\ce{C}\left(s\right)+{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{CO}}_{2}\left(g\right)\qquad\Delta{H}^{\circ }=-394\text{ kJ}[/latex]

In the two-step process, first carbon monoxide is formed:

[latex]\ce{C}\left(s\right)+\frac{1}{2}{\ce{O}}_{2}\left(g\right)\rightarrow\ce{CO}\left(g\right)\qquad\Delta{H}^{\circ }=-111\text{ kJ}[/latex]

Then, carbon monoxide reacts further to form carbon dioxide:

[latex]\ce{CO}\left(g\right)+\frac{1}{2}{\ce{O}}_{2}\left(g\right)\rightarrow\ce{CO}\left(g\right)\qquad\Delta{H}^{\circ }=-283\text{ kJ}[/latex]

The equation describing the overall reaction is the sum of these two chemical changes:

[latex]\begin{array}{l}\\ \text{Step 1: }\ce{C}\left(s\right)+\frac{1}{2}{\ce{O}}_{2}\left(g\right)\rightarrow\ce{CO}\left(g\right)\\ \underline{\text{Step 2: }\ce{CO}\left(g\right)+\frac{1}{2}{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{CO}}_{2}\left(g\right)}\\ \text{Sum: }\ce{C}\left(s\right)+\frac{1}{2}{\ce{O}}_{2}\left(g\right)+\ce{CO}\left(g\right)+\frac{1}{2}{\ce{O}}_{2}\left(g\right)\rightarrow\ce{CO}\left(g\right)+{\ce{CO}}_{2}\left(g\right)\end{array}[/latex]

Because the [latex]\ce{CO}[/latex] produced in Step 1 is consumed in Step 2, the net change is:

[latex]\ce{C}\left(s\right)+{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{CO}}_{2}\left(g\right)[/latex]

According to Hess’s law, the enthalpy change of the reaction will equal the sum of the enthalpy changes of the steps. We can apply the data from the experimental enthalpies of combustion in Table 1 to find the enthalpy change of the entire reaction from its two steps:

[latex]\begin{array}{ll}\ce{C}\left(s\right)+\frac{1}{2}{\ce{O}}_{2}\left(g\right)\rightarrow\ce{CO}\left(g\right)\hfill & \Delta{H}^{\circ }=-111\text{ kJ}\hfill \\ \underline{\ce{CO}\left(g\right)+\frac{1}{2}{\ce{O}}_{2}\left(g\right)\rightarrow{\text{CO}}_{2}\left(g\right)}& \underline{\Delta{H}^{\circ }=-283\text{ kJ}}\\{\ce{C}\left(s\right)+{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{CO}}_{2}\left(g\right)}&{\Delta{H}^{\circ }=-394\text{ kJ}}\hfill \end{array}[/latex]

The result is shown in Figure 10.3.6. We see that ΔH of the overall reaction is the same whether it occurs in one step or two. This finding (overall ΔH for the reaction = sum of ΔH values for reaction “steps” in the overall reaction) is true in general for chemical and physical processes.

A diagram is shown. A long arrow faces upward on the left with the phrase “H increasing.” A horizontal line at the bottom of the diagram is shown with the formula “C O subscript 2 (g)” below it. A horizontal line at the top of the diagram has the formulas “C (s) + O subscript 2 (g)” above it. The top and bottom lines are connected by a downward facing arrow with the value “Δ H = –394 k J” written beside it. Below and to the right of the top horizontal line is a second horizontal line with the equations “C O (g) + one half O subscript 2 (g)” above it. This line and the bottom line are connected by a downward facing arrow with the value “Δ H = –283 k J” written beside it. The same line and the top line are connected by a downward facing arrow with the value “Δ H = –111 k J” written beside it. There are three brackets to the right of the diagram. The first bracket runs from the top horizontal line to the second horizontal line. It is labeled, “Enthalpy of reactants.” The second bracket runs from the second horizontal line to the bottom horizontal line. It is labeled, “Enthalpy of products.” Both of these brackets are included in the third bracket which runs from the top to the bottom of the diagram. It is labeled, “Enthalpy change of exothermic reaction in 1 or 2 steps.”
Figure 10.3.6. The formation of [latex]\ce{CO2}[/latex](g) from its elements can be thought of as occurring in two steps, which sum to the overall reaction, as described by Hess’s law. The horizontal blue lines represent enthalpies. For an exothermic process, the products are at lower enthalpy than are the reactants.

Before we further practice using Hess’s law, let us recall two important features of ΔH.

  1. ΔH is directly proportional to the quantities of reactants or products. For example, the enthalpy change for the reaction forming 1 mole of [latex]\ce{NO2}[/latex](g) is +33.2 kJ:

    [latex]\frac{1}{2}{\ce{N}}_{2}\left(g\right)+{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{NO}}_{2}\left(g\right)\qquad\Delta\text{H}=\text{+33.2}\text{ kJ}[/latex]

    When 2 moles of [latex]\ce{NO2}[/latex] (twice as much) are formed, the ΔH will be twice as large:

    [latex]{\ce{N}}_{2}\left(g\right)+2{\ce{O}}_{2}\left(g\right)\rightarrow 2{\ce{NO}}_{2}\left(g\right)\qquad\Delta\text{H}=\text{+66.4}\text{ kJ}[/latex]

    In general, if we multiply or divide an equation by a number, then the enthalpy change should also be multiplied or divided by the same number.

  2. ΔH for a reaction in one direction is equal in magnitude and opposite in sign to ΔH for the reaction in the reverse direction. For example, given that:

    [latex]{\ce{H}}_{2}\left(g\right)+{\ce{Cl}}_{2}\left(g\right)\rightarrow 2\ce{HCl}\left(g\right)\qquad\Delta\text{H}=-184.6\text{ kJ}[/latex]

    Then, for the “reverse” reaction, the enthalpy change is also “reversed:”

    [latex]2\ce{HCl}\left(g\right)\rightarrow{\ce{H}}_{2}\left(g\right)+{\ce{Cl}}_{2}\left(g\right)\qquad\Delta\text{H}=\text{+184.6}\text{ kJ}[/latex]

Example 10.3.6: Stepwise Calculation of [latex]\Delta{H}_{\text{f}}^{\circ }[/latex] Using Hess’s Law

Determine the enthalpy of formation, [latex]\Delta{H}_{\text{f}}^{\circ }[/latex], of [latex]\ce{FeCl3}[/latex](s) from the enthalpy changes of the following two-step process that occurs under standard state conditions:

[latex]\ce{Fe}\left(s\right)+{\ce{Cl}}_{2}\left(g\right)\rightarrow{\ce{FeCl}}_{2}\left(s\right)\qquad\Delta\text{H}^{\circ }=-341.8\text{ kJ}[/latex]

[latex]{\ce{FeCl}}_{2}\left(s\right)+\frac{1}{2}{\ce{Cl}}_{2}\left(g\right)\rightarrow{\ce{FeCl}}_{3}\left(s\right)\qquad\Delta\text{H}^{\circ }=-57.7\text{ kJ}[/latex]

[reveal-answer q=”783173″]Show Solution[/reveal-answer]
[hidden-answer a=”783173″]

We are trying to find the standard enthalpy of formation of [latex]\ce{FeCl3}[/latex](s), which is equal to ΔH° for the reaction:

[latex]\ce{Fe}\left(s\right)+\frac{3}{2}{\ce{Cl}}_{2}\left(g\right)\rightarrow{\ce{FeCl}}_{3}\left(s\right)\qquad\Delta{H}_{\text{f}}^{\circ }=?[/latex]

Looking at the reactions, we see that the reaction for which we want to find ΔH° is the sum of the two reactions with known ΔH values, so we must sum their ΔHs:

[latex]\begin{array}{lll}\ce{Fe}\left(s\right)+{\ce{Cl}}_{2}\left(g\right)\rightarrow{\ce{FeCl}}_{\ce{2}}\left(s\right)\hfill & \hfill & \Delta H^{\circ }=-341.8\text{ kJ}\hfill \\ \underline{{\ce{FeCl}}_{\ce{2}}\left(s\right)+\frac{1}{2}{\text{Cl}}_{2}\left(g\right)\rightarrow{\ce{FeCl}}_{\text{3}}\left(s\right)}\hfill & \hfill & \underline{\Delta H^{\circ }=-57.7\text{ kJ}}\\{\ce{Fe}\left(s\right)+\frac{\ce{1}}{\ce{2}}{\ce{Cl}}_{2}\left(g\right)\rightarrow{\ce{FeCl}}_{\ce{3}}\left(s\right)}\hfill & \hfill & {\Delta H^{\circ }=-399.5\text{ kJ}}\hfill \end{array}[/latex]

The enthalpy of formation, [latex]\Delta{H}_{\text{f}}^{\circ }[/latex], of [latex]\ce{FeCl3}[/latex](s) is −399.5 kJ/mol.

[/hidden-answer]

Check Your Learning

Example 10.3.7: A More Challenging Problem Using Hess’s Law

Chlorine monofluoride can react with fluorine to form chlorine trifluoride:

(i) [latex]\ce{ClF}\left(g\right)+{\ce{F}}_{2}\left(g\right)\rightarrow{\ce{ClF}}_{3}\left(g\right)\qquad\Delta\text{H}^{\circ }=?[/latex]

Use the reactions here to determine the ΔH° for reaction (i):

(ii) [latex]2{\ce{OF}}_{2}\left(g\right)\rightarrow{\ce{O}}_{2}\left(g\right)+2{\ce{F}}_{2}\left(g\right)\qquad\Delta{H}_{\left(ii\right)}^{\circ }=-49.4\text{ kJ}[/latex]

(iii) [latex]2\ce{ClF}\left(g\right)+{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{Cl}}_{2}\ce{O}\left(g\right)+{\text{OF}}_{2}\left(g\right)\qquad\Delta{H}_{\left(iii\right)}^{\circ }=\text{+214.0}\text{ kJ}[/latex]

(iv) [latex]{\ce{ClF}}_{3}\left(g\right)+{\ce{O}}_{2}\left(g\right)\rightarrow\frac{1}{2}{\ce{Cl}}_{2}\ce{O}\left(g\right)+\frac{3}{2}{\ce{OF}}_{2}\left(g\right)\qquad\Delta{H}_{\left(iv\right)}^{\circ }=\text{+236.2}\text{ kJ}[/latex]

[reveal-answer q=”122087″]Show Solution[/reveal-answer]
[hidden-answer a=”122087″]

Our goal is to manipulate and combine reactions (ii), (iii), and (iv) such that they add up to reaction (i). Going from left to right in (i), we first see that [latex]\ce{ClF}[/latex](g) is needed as a reactant. This can be obtained by multiplying reaction (iii) by [latex]\frac{1}{2},[/latex] which means that the ΔH° change is also multiplied by [latex]\frac{1}{2}\text{:}[/latex]

[latex]\ce{ClF}\left(g\right)+{\ce{O}}_{2}\left(g\right)\rightarrow\frac{1}{2}{\ce{Cl}}_{2}\ce{O}\left(g\right)+\frac{1}{2}{\ce{OF}}_{2}\left(g\right)\qquad\Delta\text{H}^{\circ }=\frac{1}{2}\left(214.0\right)=\text{+107.0}\text{ kJ}[/latex]

Next, we see that [latex]\ce{F2}[/latex] is also needed as a reactant. To get this, reverse and halve reaction (ii), which means that the ΔH° changes sign and is halved:

[latex]\frac{1}{2}{\ce{O}}_{2}\left(g\right)+{\ce{F}}_{2}\left(g\right)\rightarrow{\ce{OF}}_{2}\left(g\right)\qquad\Delta\text{H}^{\circ }=\text{+24.7}\text{ kJ}[/latex]

To get [latex]\ce{ClF3}[/latex] as a product, reverse (iv), changing the sign of ΔH°:

[latex]\frac{1}{2}{\ce{Cl}}_{2}\ce{O}\left(g\right)+\frac{3}{2}{\ce{OF}}_{2}\left(g\right)\rightarrow{\ce{ClF}}_{3}\left(g\right)+{\ce{O}}_{2}\left(g\right)\qquad\Delta\text{H}^{\circ }=\text{-236.2}\text{ kJ}[/latex]

Now check to make sure that these reactions add up to the reaction we want:

[latex]\begin{array}{lll}\ce{ClF}\left(g\right)+\frac{1}{2}{\ce{O}}_{2}\left(g\right)\rightarrow\frac{1}{2}{\ce{Cl}}_{2}\ce{O}\left(g\right)+\frac{1}{2}{\ce{OF}}_{2}\left(g\right)\hfill & \hfill & \Delta H^{\circ }=\text{+107.0}\text{ kJ}\\\frac{1}{2}{\ce{O}}_{2}\left(g\right)+{\ce{F}}_{2}\left(g\right)\rightarrow{\ce{OF}}_{2}\left(g\right)\hfill & \hfill & \Delta H^{\circ }=\text{+24.7}\text{ kJ}\\\underline{\frac{1}{2}{\ce{Cl}}_{2}\ce{O}\left(g\right)+\frac{3}{2}{\ce{OF}}_{2}\left(g\right)\rightarrow{\ce{ClF}}_{3}\left(g\right)+{\ce{O}}_{2}\left(g\right)}\hfill & \hfill & \underline{\Delta H^{\circ }=-236.2\text{ kJ}}\\{\ce{ClF}\left(g\right)+{\ce{F}}_{2}\rightarrow{\ce{ClF}}_{3}\left(g\right)}\hfill & \hfill &{\Delta H^{\circ }=-104.5\text{ kJ}}\end{array}[/latex]

Reactants [latex]\frac{1}{2}{\ce{O}}_{2}[/latex] and [latex]\frac{1}{2}{\ce{O}}_{2}[/latex] cancel out product O2; product [latex]\frac{1}{2}{\ce{Cl}}_{2}\ce{O}[/latex] cancels reactant [latex]\frac{1}{2}{\ce{Cl}}_{2}\ce{O;}[/latex] and reactant [latex]\frac{3}{2}{\ce{OF}}_{2}[/latex] is cancelled by products [latex]\frac{1}{2}{\ce{OF}}_{2}[/latex] and OF2. This leaves only reactants ClF(g) and F2(g) and product ClF3(g), which are what we want. Since summing these three modified reactions yields the reaction of interest, summing the three modified ΔH° values will give the desired ΔH°:

[latex]\Delta H^{\circ }=\left(+107.0\text{ kJ}\right)+\left(24.7\text{ kJ}\right)+\left(-236.2\text{ kJ}\right)=-104.5\text{ kJ}[/latex]

[/hidden-answer]

Check Your Learning

We also can use Hess’s law to determine the enthalpy change of any reaction if the corresponding enthalpies of formation of the reactants and products are available. The stepwise reactions we consider are: (i) decompositions of the reactants into their component elements (for which the enthalpy changes are proportional to the negative of the enthalpies of formation of the reactants), followed by (ii) re-combinations of the elements to give the products (with the enthalpy changes proportional to the enthalpies of formation of the products). The standard enthalpy change of the overall reaction is therefore equal to: (ii) the sum of the standard enthalpies of formation of all the products plus (i) the sum of the negatives of the standard enthalpies of formation of the reactants. This is usually rearranged slightly to be written as follows, with ∑ representing “the sum of” and n standing for the stoichiometric coefficients:

[latex]\Delta{H}_{\text{reaction}}^{\circ }=\sum n\times \Delta{H}_{\text{f}}^{\circ }\left(\text{products}\right)-\sum n\times \Delta{H}_{\text{f}}^{\circ }\left(\text{reactants}\right)[/latex]

The following example shows in detail why this equation is valid, and how to use it to calculate the enthalpy change for a reaction of interest.

Example 10.3.8: Using Hess’s Law

What is the standard enthalpy change for the reaction:

[latex]3{\ce{NO}}_{2}\left(g\right)+{\ce{H}}_{2}\ce{O}\left(l\right)\rightarrow 2{\ce{HNO}}_{3}\left(aq\right)+\ce{NO}\left(g\right)\qquad\Delta\text{H}^{\circ }=?[/latex]

There are two possible solutions for this reaction. The first solution supports why the general equation is valid, the second solution uses the equation.
[reveal-answer q=”511660″]Show Solution 1[/reveal-answer]
[hidden-answer a=”511660″]

Solution 1 (Supporting Why the General Equation Is Valid)

We can write this reaction as the sum of the decompositions of [latex]\ce{3NO2}[/latex](g) and [latex]\ce{1H2O}[/latex](l) into their constituent elements, and the formation of [latex]\ce{2HNO3}[/latex](aq) and 1NO(g) from their constituent elements. Writing out these reactions, and noting their relationships to the [latex]\Delta{H}_{\text{f}}^{\circ }[/latex] values for these compounds (from Standard Thermodynamic Properties for Selected Substances), we have:

[latex]3{\ce{NO}}_{2}\left(g\right)\rightarrow\frac{3}{2}{\ce{N}}_{2}\left(g\right)+3{\ce{O}}_{2}\left(g\right)\qquad\Delta{H}_{1}^{\circ }=-99.6\text{ kJ}[/latex]
[latex]{\ce{H}}_{2}\ce{O}\left(l\right)\rightarrow{\ce{H}}_{2}\left(g\right)+\frac{1}{2}{\ce{O}}_{2}\left(g\right)\qquad\Delta{H}_{2}^{\circ }=\text{+285.8}\text{ kJ}\left[-1\times \Delta{H}_{\text{f}}^{\circ }\left({\ce{H}}_{2}\ce{O}\right)\right][/latex]
[latex]{\ce{H}}_{2}\left(g\right)+{\ce{N}}_{2}\left(g\right)+3{\ce{O}}_{2}\left(g\right)\rightarrow 2{\ce{HNO}}_{3}\left(aq\right)\qquad\Delta{H}_{3}^{\circ }=-414.8\text{ kJ}\left[2\times \Delta{H}_{\text{f}}^{\circ }\left({\ce{HNO}}_{3}\right)\right][/latex]
[latex]\frac{1}{2}{\ce{N}}_{2}\left(g\right)+\frac{1}{2}{\ce{O}}_{2}\left(g\right)\rightarrow\ce{NO}\left(g\right)\qquad\Delta{H}_{4}^{\circ }=\text{+90.2}\text{ kJ}\left[1\times \left(\ce{NO}\right)\right][/latex]

Summing these reaction equations gives the reaction we are interested in:

[latex]{\ce{3NO}}_{2}\left(g\right)+{\ce{H}}_{2}\ce{O}\left(l\right)\rightarrow 2{\ce{HNO}}_{3}\left(aq\right)+\ce{NO}\left(g\right)[/latex]

Summing their enthalpy changes gives the value we want to determine:

[latex]\begin{array}{cc}\hfill \Delta{H}_{\text{rxn}}^{\circ }& =\Delta{H}_{1}^{\circ }+\Delta{H}_{2}^{\circ }+\Delta{H}_{3}^{\circ }+\Delta{H}_{4}^{\circ }=\left(-99.6\text{kJ}\right)+\left(\text{+285.8}\text{kJ}\right)+\left(-414.8\text{kJ}\right)+\left(\text{+90.2}\text{kJ}\right)\hfill \\ & =-138.4\text{ kJ}\hfill \end{array}[/latex]

So the standard enthalpy change for this reaction is[latex]\Delta{H}^{\circ}=−138.4\text{ kJ}[/latex].

Note that this result was obtained by (1) multiplying the [latex]\Delta{H}_{\text{f}}^{\circ }[/latex] of each product by its stoichiometric coefficient and summing those values, (2) multiplying the [latex]\Delta{H}_{\text{f}}^{\circ }[/latex] of each reactant by its stoichiometric coefficient and summing those values, and then (3) subtracting the result found in (2) from the result found in (1). This is also the procedure in using the general equation, as shown.

[/hidden-answer]
[reveal-answer q=”565595″]Show Solution 2[/reveal-answer]
[hidden-answer a=”565595″]

Solution 2 (Using the Equation)

Alternatively, we could use the special form of Hess’s law given previously:

[latex]\begin{array}{rcl}{}\Delta{H}_{\text{reaction}}^{\circ }&=&\sum n\times \Delta{H}_{\text{f}}^{\circ }\text{(products)}-\sum n\times \Delta{H}_{\text{f}}^{\circ }\left(\text{reactants}\right)\\{}&=&\left[2\cancel{\text{mol}{\ce{HNO}}_{3}}\times \frac{-207.4\text{kJ}}{\cancel{\text{mol}{\ce{HNO}}_{3}\left(aq\right)}}+1\cancel{\text{mol}\ce{NO}\left(g\right)}\times \frac{\text{+90.2}\text{kJ}}{\cancel{\text{mol}\ce{NO}\left(g\right)}}\right]-\left[3\cancel{\text{mol}{\ce{NO}}_{2}\left(g\right)}\times \frac{\text{+33.2}\text{kJ}}{\cancel{\text{mol}{\ce{NO}}_{2}\left(g\right)}}+1\cancel{\text{mol}{\ce{H}}_{2}\ce{O}\left(l\right)}\times \frac{-285.8\text{kJ}}{\cancel{\text{mol}{\ce{H}}_{2}\ce{O}\left(l\right)}}\right]\\{}&=&2\left(-207.4\text{kJ}\right)+1\left(\text{+90.2}\text{kJ}\right)-3\left(\text{+33.2}\text{kJ}\right)-1\left(-285.8\text{kJ}\right)\\{}&=&-138.4\text{ kJ}\end{array}[/latex]

[/hidden-answer]

Check Your Learning

Key Concepts and Summary

If the enthalpies of formation are available for the reactants and products of a reaction, the enthalpy change can be calculated using Hess’s law: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps.

Key Equations

  • [latex]\Delta U=q+w[/latex]
  • [latex]\Delta{H}_{\text{reaction}}^{\circ }=\sum n\times \Delta{H}_{\text{f}}^{\circ }\text{(products)}-\sum n\times \Delta{H}_{\text{f}}^{\circ }\left(\text{reactants}\right)[/latex]

Try It

  1. Explain how the heat measured in Example 10.2.3 of Calorimetry differs from the enthalpy change for the exothermic reaction described by the following equation:
    1. [latex]\ce{HCl}\left(aq\right)+\ce{NaOH}\left(aq\right)\rightarrow\ce{NaCl}\left(aq\right)+{\ce{H}}_{2}\ce{O}\left(l\right)[/latex]
  2. Calculate the enthalpy of solution (ΔH for the dissolution) per mole of NH4NO3 under the conditions described in Figure 10.2.6 in Calorimetry.
  3. How much heat is produced by burning 4.00 moles of acetylene under standard state conditions?
  4. How many moles of isooctane must be burned to produce 100 kJ of heat under standard state conditions?
  5. When 2.50 g of methane burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion per mole of methane under these conditions?
  6. Does the standard enthalpy of formation of [latex]\ce{H2O}[/latex](g) differ from ΔH° for the reaction [latex]{\ce{2H}}_{2}\left(g\right)+{\ce{O}}_{2}\left(g\right)\rightarrow 2{\ce{H}}_{2}\ce{O}\left(g\right)?[/latex]
  7. How many kilojoules of heat will be released when exactly 1 mole of manganese, [latex]\ce{Mn}[/latex], is burned to form [latex]\ce{Mn3O4}[/latex](s) at standard state conditions?
  8. The following sequence of reactions occurs in the commercial production of aqueous nitric acid:
    1. [latex]\begin{array}{cc}4{\ce{NH}}_{3}\left(g\right)+5{\ce{O}}_{2}\left(g\right)\rightarrow 4\ce{NO}\left(g\right)+6{\ce{H}}_{2}\ce{O}\left(l\right) & \Delta\text{H}=-907\text{ kJ} \\ 2\ce{NO}\left(g\right)+{\ce{O}}_{2}\left(g\right)\rightarrow 2{\ce{NO}}_{2}\left(g\right) & \Delta\text{H}=-113\text{ kJ} \\ 3{\ce{NO}}_{2}+{\ce{H}}_{2}\ce{O}\left(l\right)\rightarrow 2{\ce{HNO}}_{2}\left(aq\right)+\ce{NO}\left(g\right) & \Delta\text{H}=-139\text{ kJ}\end{array}[/latex]
  9. Determine the total energy change for the production of one mole of aqueous nitric acid by this process.
  10. Calculate the standard molar enthalpy of formation of NO(g) from the following data:
    1. [latex]\begin{array}{ll}{\ce{N}}_{2}\left(g\right)+2{\ce{O}}_{2}\rightarrow 2{\ce{NO}}_{2}\left(g\right) & \Delta{H}_{298}^{\circ }=66.4\text{ kJ}\\ \ce{2NO}\left(g\right)+{\ce{O}}_{2}\rightarrow 2{\ce{NO}}_{2}\left(g\right) & \Delta{H}_{298}^{\circ }=-114.1\text{ kJ}\end{array}[/latex]
  11. Using the data in Standard Thermodynamic Properties for Selected Substances, calculate the standard enthalpy change for each of the following reactions:
      1. [latex]\ce{Si}\left(s\right)+2{\ce{F}}_{2}\left(g\right)\rightarrow{\ce{SiF}}_{4}\left(g\right)[/latex]
      2. [latex]2\ce{C}\left(s\right)+2{\ce{H}}_{2}\left(g\right)+{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{CH}}_{3}{\ce{CO}}_{2}\ce{H}\left(l\right)[/latex]
      3. [latex]{\ce{CH}}_{4}\left(g\right)+{\ce{N}}_{2}\left(g\right)\rightarrow\ce{HCN}\left(g\right){\ce{+NH}}_{3}\left(g\right)[/latex]
      4. [latex]{\ce{Cs}}_{2}\left(g\right)+3{\ce{Cl}}_{2}\left(g\right)\rightarrow{\ce{CCl}}_{4}\left(g\right){\ce{+S}}_{2}{\ce{Cl}}_{2}\left(g\right)[/latex]
  12. Calculate ΔH for the process [latex]{\ce{Hg}}_{2}{\ce{Cl}}_{2}\left(s\right)\rightarrow 2\ce{Hg}\left(l\right)+{\ce{Cl}}_{2}\left(g\right)[/latex] from the following information:
    1. [latex]\begin{array}{ll}\\ \ce{Hg}\left(l\right)+{\ce{Cl}}_{\ce{2}}\left(g\right)\rightarrow{\ce{HgCl}}_{\ce{2}}\left(s\right) & \Delta H=-224\text{ kJ}\\ \ce{Hg}\left(l\right)+{\ce{HgCl}}_{\ce{2}}\left(s\right)\rightarrow{\ce{Hg}}_{\ce{2}}{\ce{Cl}}_{2}\left(s\right) & \Delta H=-41.2\text{ kJ}\end{array}[/latex]
  13. Calculate the enthalpy of combustion of butane, [latex]\ce{C4H_{10}}[/latex](g) for the formation of [latex]\ce{H2O}[/latex](g) and [latex]\ce{CO2}[/latex](g). The enthalpy of formation of butane is −126 kJ/mol.
  14. The enthalpy of combustion of hard coal averages −35 kJ/g, that of gasoline, 1.28 × 105 kJ/gal. How many kilograms of hard coal provide the same amount of heat as is available from 1.0 gallon of gasoline? Assume that the density of gasoline is 0.692 g/mL (the same as the density of isooctane).

[reveal-answer q=”631741″]Show Selected Solutions[/reveal-answer]
[hidden-answer a=”631741″]

  1.  The enthalpy change of the indicated reaction is for exactly 1 mol [latex]\ce{HCL}[/latex] and 1 mol [latex]\ce{NaOH}[/latex]; the heat in the example is produced by 0.0500 mol [latex]\ce{HCl}[/latex] and 0.0500 mol [latex]\ce{NaOH}[/latex].
  2. The molar mass of [latex]\ce{NH4NO3}[/latex] is 80.0423 g/mol. From the example, 1000 J is required to dissolve 3.21 g of [latex]\ce{NH4NO3}[/latex]. One mole under the same conditions would require

    [latex]\dfrac{80.0432\cancel{\text{g}}{\text{ mol}}^{-1}}{3.21\cancel{\text{g}}}\times 1000\text{ J}=25\text{ kJ}{\text{ mol}}^{-1}.[/latex]

    (The heat of solution is positive because the process is endothermic.)

  3. The heat of combustion is -1301.1 as given in Table 10.3.1.

    heat released = 4.00 mol × (-1301.1 kJ/mol) = 5204.4 kJ

  4. The value of ΔHcomb = -5461 kJ/mol. To produce 100 kJ requires:

    [latex]\dfrac{100\text{ kJ}}{5461\text{ kJ}{\text{mol}}^{-1}}=1.83\times {10}^{-2}\text{ mol}.[/latex]

  5. The molar mass of [latex]\ce{CH4}[/latex] is 16.04 g/mol. Find the mole of [latex]\ce{CH4}[/latex] present:

    [latex]\dfrac{2.50\cancel{\text{g}}}{16.04\cancel{\text{g}}{\text{mol}}^{-1}}=0.15586\text{ mol}[/latex]

    [latex]\Delta{H}_{\text{comb}}=\dfrac{125\text{ kJ}}{0.15586\text{mol}}=802\text{ kJ}{\text{mol}}^{-1}.[/latex]

  6. No. The standard enthalpy of formation can be determined for anything, including [latex]\ce{H2O}[/latex](g), and water does not have to be liquid in this case, it’s the gas-phase water that is the substance for which the heat of formation is to be found. However, the heat of this reaction is defined for two moles of [latex]\ce{H2O}[/latex](g), thus the heat of formation is half of the heat of the reaction.
  7. This process requires 3 mol of Mn. For 1 mol, [latex]\frac{1}{3}\left(-1378.83\text{ kJ}\right)=459.6\text{ kJ}.[/latex]
  8. Enough material must be produced in each stage to proceed with the next. So [latex]\frac{1}{2}[/latex] of reaction 1 produces the [latex]\ce{NO}[/latex] required in reaction 2. But [latex]\frac{3}{2}[/latex] units of reaction 2 are required to provide enough [latex]\ce{NO2}[/latex] for reaction 3. Up to this stage, the heat produced is [latex]\frac{3}{2}\left[\frac{1}{2}\left(-907\right)+\left(-113\right)\right]=\frac{3}{2}\left(-566.55\right)=-850\text{ kJ}[/latex] to have the material to proceed with reaction 3. Therefore, from beginning to end, –850 + (–139) = –989 kJ are released. The question asks for the enthalpy change for 1 mole. Therefore, division of the last answer by 2 gives -495 kJ/mol.
  9. Hess’s law can be applied to the two equations by reversing the direction of the second equation. The first equation is a formation reaction and is so indicated by writing [latex]\Delta{H}_{298}^{\circ }[/latex].

    [latex]\begin{array}{l}{\ce{N}}_{2}\left(g\right)+2{\ce{O}}_{2}\left(g\right)\rightarrow 2{\ce{NO}}_{2}\left(g\right)\Delta{H}_{298}^{\circ }=66.4\text{ kJ}\\ 2{\ce{NO}}_{\ce{2}}\left(g\right)\rightarrow 2\ce{NO}\left(g\right)+{\ce{O}}_{2}\left(g\right)\Delta H^{\circ }=114.1\text{ kJ}\end{array}[/latex]

    Adding the equations yields:

    [latex]{\ce{N}}_{2}\left(g\right)+{\ce{O}}_{2}\left(g\right)\rightarrow 2\ce{NO}\left(g\right)\Delta{H}^{\circ }=180.5\text{ kJ}[/latex]

    This is the heat of formation of 2 mol of [latex]\ce{NO}[/latex]. For 1 mol,

    [latex]\Delta{H}_{298}^{\circ }=\frac{180.5\text{ kJ}}{2}=90.3{\text{ mol}}^{-1}\text{of}\ce{ NO}.[/latex]

  10. The standard enthalpy changes are as follows:
    1. [latex]\ce{Si}\left(s\right)+2{\ce{F}}_{2}\left(g\right)\rightarrow{\ce{SiF}}_{4}\left(g\right)[/latex]
      [latex]\begin{array}{rll} \\ \Delta{H}_{\text{reaction}}^{\circ }&=&\Delta{H}_{\text{products}}^{\circ }-\Delta{H}_{\text{reactants}}^{\circ }\\ &=&\Delta{H}_{{\ce{SiF}}_{4}\left(g\right)}^{\circ }-\Delta{H}_{\ce{Si}\left(s\right)}^{\circ }-2\Delta{H}_{{\ce{F}}_{2}\left(g\right)}^{\circ }\\ &=&-1614.9-\left(0\right)-2\left(0\right)=-1615.0\text{ kJ}{\text{mol}}^{-1}\end{array}[/latex]
    2. [latex]2\ce{C}\left(s\right)+2{\ce{H}}_{2}\left(g\right)+{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{CH}}_{3}{\ce{CO}}_{2}\ce{H}\left(l\right)[/latex]
      [latex]\begin{array}{rll} \\ \Delta{H}_{\text{reaction}}^{\circ }&=&\Delta{H}_{\text{products}}^{\circ }-\Delta{H}_{\text{reactants}}^{\circ }\\ &=&\Delta{H}_{{\ce{CH}}_{3}{\ce{CO}}_{2}\ce{H}\left(l\right)}^{\circ }-2\Delta{H}_{\ce{C}\left(s\right)}^{\circ }-2\Delta{H}_{{\ce{H}}_{2}\left(g\right)}^{\circ }-\Delta{H}_{{\ce{O}}_{2}\left(g\right)}^{\circ }\\ &=&-484.5-2\left(0\right)-2\left(0\right)-\left(0\right)=-484.3\text{ kJ}{\text{mol}}^{-1}\end{array}[/latex]
    3. [latex]{\text{CH}}_{\text{4}}\left(g\right)\rightarrow\ce{C}\left(s\right)+2{\ce{H}}_{2}\left(g\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Delta{H}_{1}^{\circ }=-\left(-74.6\text{ kJ}\right)[/latex]
      [latex]\begin{array}{lll} \\ \frac{\ce{1}}{\ce{2}}{\ce{H}}_{\ce{2}}\left(g\right)+\ce{C}\left(s\right)+\frac{1}{2}{\ce{N}}_{2}\left(g\right)\rightarrow\ce{HCN}\left(g\right)\hfill & \hfill & \Delta{H}_{2}^{\circ }=135.5\text{ kJ}\hfill \\ \underline{\frac{\text{1}}{\ce{2}}{\ce{N}}_{\ce{2}}\left(g\right)+\frac{3}{2}{\ce{H}}_{2}\left(g\right)\rightarrow{\ce{NH}}_{3}\left(g\right)}\hfill & \hfill & \underline{\Delta{H}_{3}^{\circ }=-45.9\text{ kJ}}\\{{\ce{CH}}_{4}\left(g\right)+{\ce{N}}_{2}\left(g\right)\rightarrow\ce{HCN}\left(g\right)+{\text{NH}}_{3}\left(g\right)}\hfill & \hfill &{\Delta H^{\circ }=164.2\text{ kJ}}\end{array}[/latex]
    4. [latex]{\ce{CS}}_{\ce{2}}\left(g\right)\rightarrow\ce{C}\left(s\right)+2S\left(s\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta{H}_{1}^{\circ }=-\left(116.9\text{ kJ}\right)[/latex]
      [latex]\begin{array}{lll} \\ \ce{C}\left(s\right)+2{\ce{Cl}}_{2}\left(g\right)\rightarrow{\ce{CCl}}_{\ce{4}}\left(g\right)\hfill & \hfill & \Delta{H}_{2}^{\circ }=-95.7\text{ kJ}\hfill \\ \underline{2\ce{S}\left(s\right)+{\ce{Cl}}_{2}\left(g\right)\rightarrow{\ce{S}}_{2}{\ce{Cl}}_{2}\left(g\right)}\hfill & \hfill & \underline{\Delta{H}_{3}^{\circ }=-19.50\text{ kJ}}\\{{\ce{CS}}_{2}\left(g\right)+3{\ce{Cl}}_{2}\left(g\right)\rightarrow{\ce{CCl}}_{4}\left(g\right)+{\ce{S}}_{\ce{2}}{\ce{Cl}}_{2}\left(g\right)}\hfill & \hfill &{\Delta H^{\circ }=-232.1\text{ kJ}}\end{array}[/latex]
  11. Reverse the direction of both equations and add the new equations and enthalpies.

    [latex]\begin{array}{lll}{\text{HgCl}}_{2}\rightarrow\text{Hg}\left(l\right)+{\text{Cl}}_{2}\left(l\right)\hfill & \hfill & \Delta H=224\text{ kJ}\hfill \\ \underline{{\text{Hg}}_{\text{2}}{\text{Cl}}_{2}\left(s\right)\rightarrow\text{Hg}\left(l\right)+{\text{HgCl}}_{2}\left(s\right)}\hfill & \hfill & \underline{\Delta H=41.2\text{ kJ}}\\{{\text{Hg}}_{\text{2}}{\text{Cl}}_{2}\left(s\right)\rightarrow 2\text{Hg}\left(l\right)+{\text{Cl}}_{2}\left(g\right)}\hfill & \hfill &{\Delta H=265\text{ kJ}} \end{array}[/latex]

  12. The enthalpy can be found through the following steps:

    [latex]\begin{array}{llll}\text{Step}1:\hfill & 4\left[\ce{C}\left(s\right)+{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{CO}}_{2}\left(g\right)\right]\hfill & \hfill & \Delta H^{\circ }=4\left(-394\text{ kJ}{\text{mol}}^{-1}\right)\hfill \\ \text{Step}2:\hfill & 5\left[{\ce{H}}_{\ce{2}}\left(g\right)+\frac{1}{2}{\text{O}}_{2}\left(g\right)\rightarrow{\ce{H}}_{\ce{2}}\ce{O}\left(g\right)\right]\hfill & \hfill & \Delta H^{\circ }=5\left(-242\text{ kJ}{\text{mol}}^{-1}\right)\hfill \\ \text{Step}\text{3:}\hfill & \underline{{\ce{C}}_{\ce{4}}{\ce{H}}_{10}\rightarrow 4\ce{C}\left(s\right)+5{\ce{H}}_{2}\left(g\right)}\hfill & \hfill & \underline{\Delta H^{\circ }=+126\text{ kJ}{\text{mol}}^{-1}}\hfill \\ \text{Sum:}\hfill & {\ce{C}}_{\ce{4}}{\ce{H}}_{10}+\frac{13}{2}{\ce{O}}_{2}\left(g\right)\rightarrow 4{\ce{CO}}_{2}\left(g\right)+5{\ce{H}}_{2}\ce{O}\left(l\right)\hfill & \hfill & \Delta{H}_{298}^{\circ }=-2660\text{ kJ}{\text{mol}}^{-1}\hfill \end{array}[/latex]

  13. the amount of heat produced by burning of 1.0 gallon of gasoline is:

    q = 1.0 gallon × (–1.28 × 105 kJ/gal) =–1.28 × 105 kJ
    Mass × (–35 kJ/g) =–1.28 × 105 kJ
    Mass = 3657 g or 3.7 kg

[/hidden-answer]

Glossary

Hess’s law: if a process can be represented as the sum of several steps, the enthalpy change of the process equals the sum of the enthalpy changes of the steps

 

definition

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Chemistry Fundamentals Copyright © by Dr. Julie Donnelly, Dr. Nicole Lapeyrouse, and Dr. Matthew Rex is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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