When you reverse an equation in Hess’s law, reverse the sign of its enthalpy change. When you double the equation, double the enthalpy change for the amounts in that equation. If you reverse and double the same step, both operations apply.
The 2027 A-Level H2 Chemistry syllabus (9476) includes using Hess’s law for energy cycles. This worked example starts with a target equation for two moles of carbon monoxide and shows how to report the result per mole when asked.
- Write the target equation with its state symbols.
- Apply the same reversal or multiplier to the equation and its energy change.
- Distinguish the result for the equation from the value per mole.
Worked example | Make carbon monoxide from two combustion equations
The target equation is 2C(s, graphite) + O₂(g) → 2CO(g). For this practice, use −394 kJ for C(s, graphite) + O₂(g) → CO₂(g), and −283 kJ for CO(g) + ½O₂(g) → CO₂(g). Each figure corresponds to the amounts shown in its equation.
Double the first equation to give 2C(s, graphite) + 2O₂(g) → 2CO₂(g), with −788 kJ. Reverse and double the second to give 2CO₂(g) → 2CO(g) + O₂(g), with +566 kJ.
Add the equations. Cancel the two moles of carbon dioxide and cancel one mole of oxygen from each side. The remaining equation matches the target. The net change is −788 + 566 = −222 kJ for the amounts in that target equation.
Convert to an enthalpy of formation per mole
If the question asks for the standard enthalpy of formation of carbon monoxide and the supplied values apply under standard conditions, divide by two because the target equation forms two moles of CO. Using the rounded values in this exercise gives −111 kJ mol⁻¹.
A quick check is to write the one-mole formation equation, C(s, graphite) + ½O₂(g) → CO(g). Its coefficients are half of the two-mole target, so the enthalpy change for those amounts is half as large.
Check a new pair of numbers
Suppose A → B has a change of +32 kJ and C → B has a change of −18 kJ for the amounts stated. Find the change for 2A → 2C. Double the first step to give +64 kJ. Reverse and double the second to give +36 kJ. The net result is +100 kJ.
If your sum has the right sign but the equations leave B on one side, check the coefficients again. The target reaction should be visible after cancellation.
Frequently asked questions
Does reversing and doubling cancel out?
Reversing changes the sign. Doubling changes the magnitude. The operations act on different aspects of the value.
Can I cancel the same formula in two different states?
Keep the states distinct. Changing state can involve an enthalpy change.
Official information
SEAB’s 2027 A-Level H2 Chemistry syllabus (9476), Chemical Energetics and Hess’s Law. Checked 9 November 2026.