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Hess's law

The law

Hess's law: the enthalpy change for a reaction is the same whatever route is taken from reactants to products, provided the starting and finishing conditions are the same.

Enthalpy is a property of the states, not of the journey between them.
So an enthalpy change that cannot be measured directly can be found by an indirect route built from changes that can.
This matters because many important reactions cannot be measured in a calorimeter. You cannot make carbon and hydrogen combine cleanly to give methane, but you can burn all three substances, and Hess's law does the rest.

The formation route

The most common form in the exam uses formation enthalpies, and the expression is supplied in the resource booklet:

$\Delta_rH^\circ = \Sigma\Delta_fH^\circ(\text{products}) - \Sigma\Delta_fH^\circ(\text{reactants})$

The cycle behind it goes down from the reactants to the constituent elements, then back up to the products:
- going down un-makes the reactants, which is the reverse of their formation, hence the minus
- going up makes the products, hence the plus
The elements are the common meeting point of both routes, which is why $\Delta_fH^\circ$ of an element in its standard state is defined as zero.

Using the formula

Multiply each $\Delta_fH^\circ$ by the coefficient of that substance in the balanced equation.
Add the products, add the reactants, then subtract reactants from products.
Set to zero the $\Delta_fH^\circ$ of any element in its standard state.
Getting the coefficients in is essential. A missing factor of 2 or 3 is the most common arithmetic slip on this question.

Combining equations directly

Some questions give you several equations with their enthalpy changes and ask you to build the target. Three rules cover everything:

Reverse an equation → flip the sign of its $\Delta_rH$ .
Multiply an equation by $n$ → multiply its $\Delta_rH$ by $n$ .
Add the equations → add the enthalpy changes.
Work by looking at where each substance needs to end up:
- Find a substance that appears in only one of the given equations.
- Decide whether it needs to be on the left or the right of the target, and reverse that equation if necessary.
- Scale it so the amount matches the target.
- Repeat, then check that everything not in the target cancels.

Tips and Tricks

Multiply by the coefficients. For $2\text{H}_2\text{O}$ you need $2 \times \Delta_fH^\circ(\text{H}_2\text{O})$ . Omitting the 2 is the single biggest source of lost marks on this question.

Products minus reactants, in that order. Doing it the other way round gives an answer with the right magnitude and the wrong sign, which scores nothing.

Set the elements to zero and say so in your working. Writing " $\Delta_fH^\circ(\text{O}_2) = 0$ " takes two seconds and shows the marker you understand the convention.

Watch state symbols. $\Delta_fH^\circ$ for $\text{H}_2\text{O}(l)$ and $\text{H}_2\text{O}(g)$ differ by about 44 kJ mol⁻¹, and questions deliberately mix them. If a $\Delta_{vap}H$ appears among the given data, it is there because you need to convert between the two.

When combining equations, write out the manipulated equations in full before adding them and check the cancellation on paper. An answer with no working scores nothing even if the number is right — the specification requires working to be shown.

Sanity check your answer. Combustion reactions are always exothermic, so a positive $\Delta_cH$ means a sign error somewhere.

Worked ExampleTwo Hess's law routes

(a) Calculate $\Delta_rH^\circ$ for the complete combustion of ethanol: $\text{C}_2\text{H}_5\text{OH}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)$ given $\Delta_fH^\circ$ : $\text{C}_2\text{H}_5\text{OH}(l) = -277.6$ , $\text{CO}_2(g) = -393.5$ , $\text{H}_2\text{O}(l) = -285.8$ kJ mol⁻¹.

(b) Use the data below to calculate $\Delta_fH^\circ$ for methane: $\text{C}(s) + 2\text{H}_2(g) \rightarrow \text{CH}_4(g)$

	Equation	$\Delta_rH^\circ$ / kJ mol⁻¹
(1)	$\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g)$	−393.5
(2)	$\text{H}_2(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l)$	−285.8
(3)	$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$	−890.3

Part (a) — the formation route

Step 1 — Note the element. Oxygen appears as $\text{O}_2(g)$ , which is an element in its standard state, so

$\Delta_fH^\circ(\text{O}_2) = 0$

Step 2 — Sum the products, using the coefficients. There are 2 mol of $\text{CO}_2$ and 3 mol of $\text{H}_2\text{O}$ :

$\Sigma\Delta_fH^\circ(\text{products}) = 2(-393.5) + 3(-285.8)$ $= -787.0 + (-857.4) = -1644.4 \text{ kJ mol}^{-1}$

Step 3 — Sum the reactants. There is 1 mol of ethanol and 3 mol of oxygen, and the oxygen contributes nothing:

$\Sigma\Delta_fH^\circ(\text{reactants}) = -277.6 + 3(0) = -277.6 \text{ kJ mol}^{-1}$

Step 4 — Subtract, products first.

$\Delta_rH^\circ = -1644.4 - (-277.6) = -1644.4 + 277.6 = -1366.8 \text{ kJ mol}^{-1}$

Part (b) — combining equations

Step 1 — Look at where each substance must end up. The target is $\text{C}(s) + 2\text{H}_2(g) \rightarrow \text{CH}_4(g)$ .

C(s) must be on the left. In equation (1) it is already on the left → use (1) as written.
H₂(g) must be on the left, and we need 2 mol. In equation (2) it is on the left but there is only 1 mol → multiply (2) by 2.
CH₄(g) must be on the right. In equation (3) it is on the left → reverse (3), which flips the sign.

Step 2 — Write out the manipulated equations.

Equation (1), unchanged: $\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \qquad \Delta H = -393.5$

Equation (2) × 2 — double the enthalpy change too: $2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l) \qquad \Delta H = 2(-285.8) = -571.6$

Equation (3) reversed — flip the sign: $\text{CO}_2(g) + 2\text{H}_2\text{O}(l) \rightarrow \text{CH}_4(g) + 2\text{O}_2(g) \qquad \Delta H = +890.3$

Step 3 — Add them and check the cancellation. On the left we have $\text{C}(s)$ , $\text{O}_2$ , $2\text{H}_2$ , $\text{O}_2$ , $\text{CO}_2$ , $2\text{H}_2\text{O}$ . On the right we have $\text{CO}_2$ , $2\text{H}_2\text{O}$ , $\text{CH}_4$ , $2\text{O}_2$ .

Cancelling what appears on both sides: $\text{CO}_2$ cancels, $2\text{H}_2\text{O}$ cancels, and $2\text{O}_2$ cancels against the two single $\text{O}_2$ on the left.

$\text{C}(s) + 2\text{H}_2(g) \rightarrow \text{CH}_4(g) \quad \checkmark \text{ the target}$

Step 4 — Add the enthalpy changes.

$\Delta_fH^\circ(\text{CH}_4) = -393.5 + (-571.6) + 890.3$ $= -965.1 + 890.3 = -74.8 \text{ kJ mol}^{-1}$