Tables of reaction enthalpies would need one entry for every reaction anyone might run, and there are millions. Chemists instead keep one number for each substance: its standard enthalpy of formation. From a short table of these numbers you can find $ΔH$ for almost any reaction, without a calorimeter and without hunting for the right equations to combine.

In this lesson you will look up enthalpies of formation, add them for the products and for the reactants, and subtract (Figure 1). It is Hess’s law again, with every reaction routed through the elements.

1. Balanced equationwith the state of each substance
2. look up ΔH⁰f; elements in standard states are 0
3. Sums for each siden × ΔH⁰f, added
4. products minus reactants
5. ΔH⁰ of the reactionround to the least precise place
6. $\times \dfrac{n}{n_\text{equation}}$scale to the amount
7. ΔH for the amountsign kept

$\begin{aligned} &ΔH^0_\mathrm{rxn} = Σ n\,ΔH^0_\mathrm{f}(\text{products}) \\ &\ \ \ \ -Σ n\,ΔH^0_\mathrm{f}(\text{reactants}) \end{aligned}$$ΔH^0_\mathrm{f}(\text{element}) = 0$.

Learning Objectives

After this lesson, you should be able to:

1. Calculate a standard enthalpy of reaction from tabulated enthalpies of formation.

Key Ideas

1. What a Standard Enthalpy of Formation Is

The standard enthalpy of formation of a substance, $ΔH^0_\mathrm{f}$, is the enthalpy change when 1 mol of the substance forms from its elements, with everything in its standard state: the most stable form at 1 atm and a stated temperature, usually 25 °C. The superscript 0 means standard conditions. For ammonia, the formation equation makes exactly 1 mol of product, so fractions of a mole of the elements are allowed:

Forming ammonia releases heat, so its $ΔH^0_\mathrm{f}$ is negative. Forming nitrogen monoxide absorbs heat, so its value is positive:

An element already in its standard state needs no reaction to form it, so its $ΔH^0_\mathrm{f}$ is exactly zero: $\mathrm{O_2\!(g)}$, $\mathrm{N_2\!(g)}$, $\mathrm{H_2\!(g)}$, $\mathrm{Cl_2\!(g)}$, $\mathrm{Fe(s)}$ and $\mathrm{C(graphite)}$ all have $ΔH^0_\mathrm{f} = 0$. The elements make a common zero level for every compound (Figure 2): Panel (a) shows ammonia below it and Panel (b) shows nitrogen monoxide above it.

Key idea: $ΔH^0_\mathrm{f}$ is $ΔH$ for making 1 mol of a substance from its elements in their standard states. Elements in their standard states have $ΔH^0_\mathrm{f} = 0$; a negative value means the compound lies below its elements in enthalpy.

2. Products Minus Reactants

Any reaction can be carried out, on paper, in two steps through the elements: first break the reactants down into their elements, then build the products from those same elements. Breaking a compound down is the reverse of forming it, so that step has $ΔH = -ΔH^0_\mathrm{f}$ for each mole of reactant. Building the products has $ΔH = +ΔH^0_\mathrm{f}$ for each mole of product. By Hess’s law the two steps add up to the reaction:

Here $n$ is the coefficient of each substance in the balanced equation. Take the water-gas shift reaction, used to make hydrogen:

Hydrogen is an element, so it adds 0. The products sum is −393.5 kJ and the reactants sum is −110.5 kJ + (−241.8 kJ) = −352.3 kJ:

Figure 3 shows the same calculation as a path: up 352.3 kJ from the reactants to the elements, then down 393.5 kJ to the products, a net change of −41.2 kJ.

Key idea: $ΔH^0_\mathrm{rxn}$ is the sum for the products minus the sum for the reactants, each $ΔH^0_\mathrm{f}$ times its coefficient. Always products minus reactants; elements in their standard states add 0.

3. Coefficients and States Matter

$ΔH^0_\mathrm{f}$ is given for one mole of the substance (units kJ/mol), so multiply each value by the coefficient (in mol) of that substance before you add: 4 mol of nitrogen monoxide contributes $4 \times 90.25 = 361.00\,\text{kJ}$ (Panel (b) of Figure 4).

The state also matters. Liquid water and water vapor have different values, $ΔH^0_\mathrm{f}[\mathrm{H_2\!O(l)}] = -285.8\ \text{kJ/mol}$ and $ΔH^0_\mathrm{f}[\mathrm{H_2\!O(g)}] = -241.8\ \text{kJ/mol}$; the gas lies 44.0 kJ/mol higher because it takes energy to evaporate water (Panel (a) of Figure 4). Read the state in the equation, then pick the matching row of the table.

Key idea: multiply each $ΔH^0_\mathrm{f}$ by its coefficient, and use the value for the state written in the equation, (g), (l) or (s).

4. A Strategy for Formation-Enthalpy Problems

Follow the roadmap in Figure 1:

1. Check the equation is balanced and note the state of each substance.
2. Look up $ΔH^0_\mathrm{f}$ for each compound in that state; write 0 for each element in its standard state.
3. Add up $n\,ΔH^0_\mathrm{f}$ for the products, then for the reactants.
4. Subtract: products minus reactants. Round $ΔH^0_\mathrm{rxn}$ to the decimal place of the least precise $ΔH^0_\mathrm{f}$ you used.
5. Scale $ΔH^0_\mathrm{rxn}$, which goes with the moles in the coefficients, to the amount in the question (a mass to moles first).

Exam tip: write every $ΔH^0_\mathrm{f}$ with its sign and put negative values in parentheses before you multiply or subtract. Subtracting a negative reactants sum is where most sign errors happen.

Worked Example 1: All Coefficients 1

Problem: Ammonia gas and hydrogen chloride gas meet and form a white smoke of solid ammonium chloride. Use the standard enthalpies of formation to find $ΔH$, in kJ, when 0.420 mol of ammonia reacts.

Step 1. Add up $n\,ΔH^0_\mathrm{f}$ for the products. The only product is 1 mol of ammonium chloride.

Step 2. Add up $n\,ΔH^0_\mathrm{f}$ for the reactants.

Step 3. Subtract: products minus reactants. The least precise value, −314.4 kJ/mol, has 1 decimal place, so round to 1 decimal place.

Step 4. Scale $ΔH$ to the amount. $ΔH^0_\mathrm{rxn}$ goes with 1 mol of $\mathrm{NH_3}$, its coefficient in the equation.

Final Answer: ΔH = −73.9 kJ

The reaction is exothermic: the product lies lower in enthalpy than the two gases. Nothing had to be reversed or combined by hand; the table and one subtraction did the work.

1. Sums for each side$\begin{aligned} &\text{products: }{-314.4}\,\text{kJ} \\ &\text{reactants: }{-138.42}\,\text{kJ} \end{aligned}$
2. Steps 1 to 3: products minus reactants
3. ΔH⁰ of the reaction$ΔH^0_\mathrm{rxn} = -176.0\,\text{kJ}$Negative: exothermic
4. $\times \dfrac{0.420\,\text{mol}\;\mathrm{NH_3}}{1\,\text{mol}\;\mathrm{NH_3}}$Step 4: scale to the amount
5. ΔH for this amount$ΔH = -73.92\,\text{kJ} \approx -73.9\,\text{kJ}$

Your turn: all coefficients 1

Same four steps as Worked Example 1: products sum, reactants sum, subtract, then scale to the amount in the question. Type a plus or minus sign with your answer.

Worked Example 2: Coefficients Other Than 1

Problem: In the first step of making nitric acid, ammonia burns over a platinum catalyst. Use the standard enthalpies of formation to find $ΔH$, in kJ, when 0.850 mol of ammonia reacts.

Step 1. Add up $n\,ΔH^0_\mathrm{f}$ for the products. Multiply each value by its coefficient. The water is a gas here, so use the value for $\mathrm{H_2\!O(g)}$.

Step 2. Add up $n\,ΔH^0_\mathrm{f}$ for the reactants. Oxygen is an element in its standard state, so it adds 0.

Step 3. Subtract: products minus reactants. The least precise value, −241.8 kJ/mol, has 1 decimal place, so round to 1 decimal place.

Step 4. Scale $ΔH$ to the amount. $ΔH^0_\mathrm{rxn}$ goes with 4 mol of $\mathrm{NH_3}$.

Final Answer: ΔH = −192 kJ

$ΔH^0_\mathrm{rxn} = -905.4\ \text{kJ}$ is for the equation as written, 4 mol of ammonia, so 0.850 mol releases less than a quarter of it.

1. Sums for each side$\begin{aligned} &\text{products: }{-1089.80}\,\text{kJ} \\ &\text{reactants: }{-184.44}\,\text{kJ} \end{aligned}$each value times its coefficient
2. Steps 1 to 3: products minus reactants
3. ΔH⁰ of the reaction$ΔH^0_\mathrm{rxn} = -905.4\,\text{kJ}$for 4 mol of NH₃
4. $\times \dfrac{0.850\,\text{mol}\;\mathrm{NH_3}}{4\,\text{mol}\;\mathrm{NH_3}}$Step 4: scale to the amount
5. ΔH for this amount$ΔH = -192.4\,\text{kJ} \approx -192\,\text{kJ}$

Your turn: coefficients other than 1

Same four steps as Worked Example 2: multiply each value by its coefficient, watch the state of water, subtract, then scale. Type a plus or minus sign with your answer.

Worked Example 3: ΔH for a Mass

Problem: In a blast furnace, carbon monoxide reduces iron(III) oxide to iron. Use the standard enthalpies of formation to find $ΔH$, in kJ, when 45.6 g of iron(III) oxide reacts. The molar mass of iron(III) oxide is 159.69 g/mol.

Step 1. Add up $n\,ΔH^0_\mathrm{f}$ for the products. Iron is an element in its standard state, so it adds 0.

Step 2. Add up $n\,ΔH^0_\mathrm{f}$ for the reactants.

Step 3. Subtract: products minus reactants.

Step 4. Convert the mass of iron(III) oxide to moles.

Step 5. Scale $ΔH$ to the amount. $ΔH^0_\mathrm{rxn}$ goes with 1 mol of $\mathrm{Fe_2\!O_3}$.

Final Answer: ΔH = −7.08 kJ

Two large sums, about 1200 kJ each, differ by only a little, so this reaction releases little heat. Keeping every digit until the subtraction is what makes the small difference reliable.

1. ΔH⁰ of the reaction$ΔH^0_\mathrm{rxn} = -24.8\,\text{kJ}$products minus reactants
2. Steps 1 to 3: for each mole of Fe₂O₃
3. Mass of Fe₂O₃$45.6\,\text{g}$
4. $\times \dfrac{1\,\text{mol}}{159.69\,\text{g}}$Step 4: grams to moles
5. Moles of Fe₂O₃$0.2856\,\text{mol}\;\mathrm{Fe_2\!O_3}$
6. $\times \dfrac{-24.8\,\text{kJ}}{1\,\text{mol}\;\mathrm{Fe_2\!O_3}}$Step 5: scale to the amount
7. ΔH for this amount$ΔH = -7.083\,\text{kJ} \approx -7.08\,\text{kJ}$Negative: exothermic

Your turn: ΔH for a mass

Same five steps as Worked Example 3: products sum, reactants sum, subtract, convert grams to moles with the molar mass, then scale. Type a plus or minus sign with your answer.

Common Mistakes

Throughout, the sums are exact; keep 4 significant figures in the scaled values and round only the final answer to 3.

Common mistake: Reactants minus products. In Worked Example 1, subtracting the other way round gives $+176.0\ \text{kJ}$ instead of $-176.0\ \text{kJ}$: the right size with the wrong sign. It is always products minus reactants.

Common mistake: Adding the two sums. In Worked Example 1, adding the products sum and the reactants sum gives $-452.82\ \text{kJ}$, which is not $ΔH$ of anything. Subtract the reactants sum.

Common mistake: Leaving out the coefficients. In Worked Example 2, using each $ΔH^0_\mathrm{f}$ only once gives $-105.4\ \text{kJ}$ instead of $-905.4\ \text{kJ}$. Multiply every value by its coefficient.

Common mistake: Using the wrong state of water. In Worked Example 2 the water is a gas. Using the liquid value, −285.8 kJ/mol, gives $-1169.4\ \text{kJ}$ instead of $-905.4\ \text{kJ}$.

Common mistake: Giving ΔH of the equation instead of scaling it. In Worked Example 3, $-24.8\ \text{kJ}$ is for 1 mol of iron(III) oxide; 45.6 g is only 0.2856 mol, so $ΔH = -7.08\ \text{kJ}$.

Key Takeaway

A standard enthalpy of formation is $ΔH$ for making 1 mol of a substance from its elements in their standard states, and it is zero for an element in its standard state. Routing a reaction through the elements (Hess’s law) gives $ΔH^0_\mathrm{rxn}$ as the products sum minus the reactants sum of $n\,ΔH^0_\mathrm{f}$. Use the value for the right state, multiply by the coefficients, round to the least precise decimal place, then scale to the amount.

Practice: randomized problems

Each problem below uses fresh numbers every time and mixes every kind of problem in this lesson: work it out, check your answer, then open the worked solution to compare your steps.

Your results

This summary collects every practice problem you try on this page and suggests what to review next.

Figures: original diagrams by Dr. Ray’s Learning Center, licensed CC BY 4.0. Data: Whitten, Davis, Peck and Stanley, Chemistry, 9th ed. (Brooks/Cole, Cengage Learning, 2010), Appendix K (pp. A-23 to A-25) for the standard enthalpies of formation; Sections 15-7 and 15-8 (pp. 559 to 560 and 563 to 564) for the method; molar masses from IUPAC atomic weights, as in the practice engine. The practice problems are generated and checked by an audited engine; see the lesson audit log.