Chemistry » Thermochemistry » Enthalpy

Hess’s Law

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:

\(\text{C}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{298}^{°}=-394\phantom{\rule{0.2em}{0ex}}\text{kJ}\)

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

\(\text{C}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CO}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{298}^{°}=-111\phantom{\rule{0.2em}{0ex}}\text{kJ}\)

Then, carbon monoxide reacts further to form carbon dioxide:

\(\text{CO}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{298}^{°}=-283\phantom{\rule{0.2em}{0ex}}\text{kJ}\)

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

\(\begin{array}{}\\ \text{Step 1: C}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CO}\left(g\right)\\ \underset{¯}{\text{Step 2: CO}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)}\\ \text{Sum: C}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}{\text{O}}_{2}\left(g\right)+\text{CO}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CO}\left(g\right)+{\text{CO}}_{2}\left(g\right)\end{array}\)

Because the CO produced in Step 1 is consumed in Step 2, the net change is:

\(\text{C}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)\)

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 the table of Standard Molar Enthalpies of Combustion from the lesson on Enthalpy of Combustion to find the enthalpy change of the entire reaction from its two steps:

\(\begin{array}{ll}\text{C}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{CO}\left(g\right)\hfill & \text{Δ}{H}_{298}^{°}=-111\phantom{\rule{0.2em}{0ex}}\text{kJ}\hfill \\ \cfrac{\text{CO}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)}{\text{C}\left(s\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}\left(g\right)\phantom{\rule{1em}{0ex}}}\hfill & \cfrac{\text{Δ}{H}_{298}^{°}=-283\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\text{Δ}{H}_{298}^{°}=-394\phantom{\rule{0.2em}{0ex}}\text{kJ}}\hfill \end{array}\)

The result is shown in the figure below. 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.”

The formation of CO2(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 NO2(g) is +33.2 kJ:

    \(\cfrac{1}{2}{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}=\text{+33.2 kJ}\)

    When 2 moles of NO2 (twice as much) are formed, the ΔH will be twice as large:

    \({\text{N}}_{2}\left(g\right)+2{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{NO}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}=\text{+66.4 kJ}\)

    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:

    \({\text{H}}_{2}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2\text{HCl}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}=-184.6\phantom{\rule{0.2em}{0ex}}\text{kJ}\)

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

    \(2\text{HCl}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}=\text{+184.6 kJ}\)

Example: Stepwise Calculation of \(\text{Δ}{H}_{\text{f}}^{°}\) Using Hess’s Law

Determine the enthalpy of formation, \(\text{Δ}{H}_{\text{f}}^{°},\)of FeCl3(s) from the enthalpy changes of the following two-step process that occurs under standard state conditions:

\(\text{Fe}\left(s\right)+{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{2}\left(s\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}\text{°}=-341.8\phantom{\rule{0.2em}{0ex}}\text{kJ}\)

\({\text{FeCl}}_{2}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{3}\left(s\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}\text{°}=-57.7\phantom{\rule{0.2em}{0ex}}\text{kJ}\)

Solution

We are trying to find the standard enthalpy of formation of FeCl3(s), which is equal to ΔH° for the reaction:

\(\text{Fe}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{3}{2}{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{3}\left(s\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}=?\)

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:

\(\begin{array}{lll}\text{Fe}\left(s\right)+{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{2}\left(s\right)\hfill & \hfill & \text{Δ}H\text{°}=-341.8\phantom{\rule{0.2em}{0ex}}\text{kJ}\hfill \\ \cfrac{{\text{FeCl}}_{2}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{3}\left(s\right)}{\text{Fe}\left(s\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}{\text{Cl}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{3}\left(s\right)\phantom{\rule{1em}{0ex}}}\hfill & \hfill & \cfrac{\text{Δ}H\text{°}=-57.7\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\text{Δ}H\text{°}=-399.5\phantom{\rule{0.2em}{0ex}}\text{kJ}}\hfill \end{array}\)

The enthalpy of formation, \(\text{Δ}{H}_{\text{f}}^{°},\) of FeCl3(s) is −399.5 kJ/mol.

Here is a less straightforward example that illustrates the thought process involved in solving many Hess’s law problems. It shows how we can find many standard enthalpies of formation (and other values of ΔH) if they are difficult to determine experimentally.

Example: A More Challenging Problem Using Hess’s Law

Chlorine monofluoride can react with fluorine to form chlorine trifluoride:

(i)\(\text{ClF}\left(g\right)+{\text{F}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{ClF}}_{3}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}\text{°}=?\)

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

(ii)\(2{\text{OF}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{O}}_{2}\left(g\right)+2{\text{F}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{\left(ii\right)}^{°}=-49.4\phantom{\rule{0.2em}{0ex}}\text{kJ}\)

(iii)\(2\text{ClF}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{Cl}}_{2}\text{O}\left(g\right)+{\text{OF}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{\left(iii\right)}^{°}=\text{+205.6 kJ}\)

(iv)\({\text{ClF}}_{3}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\cfrac{1}{2}{\text{Cl}}_{2}\text{O}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{3}{2}{\text{OF}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{\left(iv\right)}^{°}=\text{+266.7 kJ}\)

Solution

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 ClF(g) is needed as a reactant. This can be obtained by multiplying reaction (iii) by \(\cfrac{1}{2},\) which means that the ΔH° change is also multiplied by \(\cfrac{1}{2}\text{:}\)

\(\text{ClF}\left(g\right)+\cfrac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\cfrac{1}{2}{\text{Cl}}_{2}\text{O}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}{\text{OF}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}\text{°}=\phantom{\rule{0.2em}{0ex}}\cfrac{1}{2}\left(205.6\right)=\text{+102.8 kJ}\)

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

\(\cfrac{1}{2}{\text{O}}_{2}\left(g\right)+{\text{F}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{OF}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}\text{°}=\text{+24.7 kJ}\)

To get ClF3 as a product, reverse (iv), changing the sign of ΔH°:

\(\cfrac{1}{2}{\text{Cl}}_{2}\text{O}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{3}{2}{\text{OF}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{ClF}}_{3}\left(g\right)+{\text{O}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}\text{°}=\text{−266.7 kJ}\)

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

\(\begin{array}{lll}\text{ClF}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\cfrac{1}{2}{\text{Cl}}_{2}\text{O}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}{\text{OF}}_{2}\left(g\right)\hfill & \hfill & \text{Δ}H\text{°}=\text{+102.8 kJ}\hfill \\ \cfrac{1}{2}{\text{O}}_{2}\left(g\right)+{\text{F}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{OF}}_{2}\left(g\right)\hfill & \hfill & \text{Δ}H\text{°}=\text{+24.7 kJ}\hfill \\ \cfrac{\cfrac{1}{2}{\text{Cl}}_{2}\text{O}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{3}{2}{\text{OF}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{ClF}}_{3}\left(g\right)+{\text{O}}_{2}\left(g\right)}{\text{ClF}\left(g\right)+{\text{F}}_{2}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{ClF}}_{3}\left(g\right)\phantom{\rule{8em}{0ex}}}\hfill & \hfill & \cfrac{\text{Δ}H\text{°}=-266.7\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\text{Δ}H\text{°}=-139.2\phantom{\rule{0.2em}{0ex}}\text{kJ}}\hfill \end{array}\)

Reactants \(\cfrac{1}{2}{\text{O}}_{2}\)and \(\cfrac{1}{2}{\text{O}}_{2}\)cancel out product O2; product \(\cfrac{1}{2}{\text{Cl}}_{2}\text{O}\) cancels reactant \(\cfrac{1}{2}{\text{Cl}}_{2}\text{O;}\) and reactant \(\cfrac{3}{2}{\text{OF}}_{2}\) is cancelled by products \(\cfrac{1}{2}{\text{OF}}_{2}\) 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°:

\(\text{Δ}H\text{°}=\left(+102.8\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)+\left(24.7\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)+\left(-266.7\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)=-139.2\phantom{\rule{0.2em}{0ex}}\text{kJ}\)

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:

\(\text{Δ}{H}_{\text{reaction}}^{°}=\sum n\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\left(\text{products}\right)-\sum n\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\left(\text{reactants}\right)\)

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: Using Hess’s Law

What is the standard enthalpy change for the reaction:

\(3{\text{NO}}_{2}\left(g\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{HNO}}_{3}\left(aq\right)+\text{NO}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}\text{H}\text{°}=?\)

Solution:  Using the Equation

Use the special form of Hess’s law given previously, and values from this lesson:

\(\require{cancel}\text{Δ}{H}_{\text{reaction}}^{°}=\sum n\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\text{(products)}-\sum n\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\left(\text{reactants}\right)\)

\(\begin{array}{l} =\left[2\phantom{\rule{0.2em}{0ex}}\cancel{\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{HNO}}_{3}\left(aq\right)}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{-207.4\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\cancel{\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{HNO}}_{3}\left(aq\right)}}\phantom{\rule{0.1em}{0ex}}+1\phantom{\rule{0.2em}{0ex}}\cancel{\text{mol NO}\left(g\right)}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{\text{+90.2 kJ}}{\cancel{\text{mol NO}\left(g\right)}}\right]\\ -\left[3\phantom{\rule{0.2em}{0ex}}\cancel{\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}\left(g\right)}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{\text{+33.2 kJ}}{\cancel{\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}\left(g\right)}}\phantom{\rule{0.1em}{0ex}}+1\phantom{\rule{0.2em}{0ex}}\cancel{\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(l\right)}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{-285.8\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\cancel{\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(l\right)}}\right]\\ =2\left(-207.4\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)+1\left(\text{+90.2 kJ}\right)-3\left(\text{+33.2 kJ}\right)-1\left(-285.8\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)\\ =-138.4\phantom{\rule{0.2em}{0ex}}\text{kJ}\end{array}\)

Solution: Supporting Why the General Equation Is ValidAlternatively, we can write this reaction as the sum of the decompositions of 3NO2(g) and 1H2O(l) into their constituent elements, and the formation of 2HNO3(aq) and 1NO(g) from their constituent elements. Writing out these reactions, and noting their relationships to the \(\text{Δ}{H}_{\text{f}}^{°}\) values for these compounds (from this table), we have:

\(3{\text{NO}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{3/2N}}_{2}\left(g\right)+{\text{3O}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{1}^{°}=-99.6\phantom{\rule{0.2em}{0ex}}\text{kJ}\)

\({\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{2}^{°}=\text{+285.8 kJ}\phantom{\rule{0.2em}{0ex}}\left[-1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\left({\text{H}}_{2}\text{O}\right)\right]\)

\({\text{H}}_{2}\left(g\right)+{\text{N}}_{2}\left(g\right)+3{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{HNO}}_{3}\left(aq\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{3}^{°}=-414.8\phantom{\rule{0.2em}{0ex}}\text{kJ}\phantom{\rule{0.2em}{0ex}}\left[2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{Δ}{H}_{\text{f}}^{°}\left({\text{HNO}}_{3}\right)\right]\)

\(\cfrac{1}{2}{\text{N}}_{2}\left(g\right)+\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}{\text{O}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}\text{NO}\left(g\right)\phantom{\rule{3em}{0ex}}\text{Δ}{H}_{4}^{°}=\text{+90.2 kJ}\phantom{\rule{0.2em}{0ex}}\left[1\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\text{NO}\right)\right]\)

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

\({\text{3NO}}_{2}\left(g\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{HNO}}_{3}\left(aq\right)+\text{NO}\left(g\right)\)

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

\(\begin{array}{cc}\hfill \text{Δ}{H}_{\text{rxn}}^{°}& =\text{Δ}{H}_{1}^{°}+\text{Δ}{H}_{2}^{°}+\text{Δ}{H}_{3}^{°}+\text{Δ}{H}_{4}^{°}=\left(-99.6\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)+\left(\text{+285.8 kJ}\right)+\left(-414.8\phantom{\rule{0.2em}{0ex}}\text{kJ}\right)+\left(\text{+90.2 kJ}\right)\hfill \\ & =-138.4\phantom{\rule{0.2em}{0ex}}\text{kJ}\hfill \end{array}\)

So the standard enthalpy change for this reaction is ΔH° = −138.4 kJ.

Note that this result was obtained by (1) multiplying the \(\text{Δ}{H}_{\text{f}}^{°}\) of each product by its stoichiometric coefficient and summing those values, (2) multiplying the \(\text{Δ}{H}_{\text{f}}^{°}\) 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.

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