Hess's Law by Equation Reversal

Introduction

Sometimes the desired reaction enthalpy is not tabulated, but enthalpies for related reactions are. Hess's law lets us combine them because enthalpy depends on initial and final states, not on the imagined route between them. The numerical challenge is disciplined equation algebra: reverse an equation only if its sign is reversed, multiply every coefficient and its ΔH by the same number, then add and cancel species.

Core explanation

If A → B has enthalpy change ΔH, the reverse B → A has −ΔH. Reversing swaps initial and final states, so the enthalpy difference changes sign. If an equation is multiplied by k, including fractions, its enthalpy change becomes kΔH because k times as much reaction is represented. When manipulated equations are added, their enthalpy changes add. Species that appear on both sides in equal amounts cancel, leaving the target equation.

For a concrete cycle, use illustrative standard data: C(graphite) + O₂(g) → CO₂(g), ΔH₁ = −393.5 kJ, and CO(g) + ½O₂(g) → CO₂(g), ΔH₂ = −283.0 kJ. The target is C(graphite) + ½O₂(g) → CO(g). Reverse the second equation: CO₂(g) → CO(g) + ½O₂(g), ΔH = +283.0 kJ. Add it to the first: CO₂ cancels, and one O₂ on the left partly cancels half O₂ on the right. The net equation is exactly the target, with ΔH target = −393.5 + 283.0 = −110.5 kJ per mole of target reaction.

The sign and cancellation should be checked separately. One could accidentally obtain −676.5 kJ by reversing the equation but keeping its original negative enthalpy. Another mistake is to cancel a species with different phases, such as H₂O(l) against H₂O(g); those are not identical thermodynamic states. Likewise, graphite and diamond cannot be cancelled as though they were the same carbon state. Include states in every thermochemical equation.

An efficient method is to find a species that appears only in one supplied equation and is needed in the target; orient and scale that equation to match its target side and coefficient. Repeat for another distinctive species, then add the equations and inspect unwanted intermediates. If an unwanted intermediate remains, another equation may be required, or a chosen orientation or multiplier may be wrong. Do not manipulate enthalpy numbers independently of chemical equations.

Hess's law is about state-function bookkeeping. The laboratory reaction may not proceed by the listed steps, and the cycle does not prove any microscopic mechanism. It also does not say a reaction is fast or spontaneous under all conditions. It supplies a thermodynamic enthalpy difference for matched initial and final states. OpenStax Chemistry 2e's enthalpy section develops the same step-addition principle.

Step-by-step reasoning

1. Write the target equation clearly with coefficients and physical states. 2. Select a supplied equation containing a target species and orient it so that species appears on the correct side. 3. Multiply its entire equation and ΔH by the factor needed for the target coefficient. 4. Repeat for other equations, add left and right sides, and cancel identical states that appear on both. 5. Confirm the net equation exactly matches the target before summing the manipulated ΔH values.

Visual explanation

Draw a triangular path with carbon plus oxygen at one corner, carbon dioxide at the second, and carbon monoxide plus half oxygen at the third. One arrow is carbon combustion to CO₂, and the second reversed arrow goes from CO₂ to CO plus oxygen. The direct arrow between start and finish is the target enthalpy, equal to the sum along the two-step route.

Real-world analogy

Travelling from town A to town C through town B has a total elevation change equal to the A-to-B change plus the B-to-C change. Walking a route backward reverses the sign of its elevation change. Taking two identical trips doubles the total change. Enthalpy cycles follow the same signed-path arithmetic.

Real-world example

Combustion measurements can be easier to obtain than the enthalpy of forming an intermediate from elements. By measuring complete combustion of an element and combustion of the intermediate, a chemist can subtract the two paths to estimate the intermediate's formation enthalpy without directly measuring the target formation reaction.

Why?

Why can different paths be added? Enthalpy is a state function. Every intermediate species introduced and then consumed has no net contribution to the overall initial and final states. Their stepwise enthalpy changes cancel through the sum, leaving the target change.

Common misconception

“Reverse the arrow but keep the reported enthalpy.” Reversal always changes the sign. Scaling an equation also scales ΔH, including when the factor is one-half. A useful final check is to verify species cancellation before touching the enthalpy arithmetic.

Worked example

Given A → B with ΔH = +40 kJ and A → C with ΔH = −10 kJ, find ΔH for B → C. Reverse A → B to B → A with ΔH = −40 kJ. Add B → A and A → C. A appears once on each side and cancels, leaving B → C. The target enthalpy is −40 + (−10) = −50 kJ per mole of the displayed B → C reaction. If the target were 2B → 2C, multiply the net equation and ΔH by two to obtain −100 kJ.

Quick check

1. A → B has ΔH = −25 kJ. What are the equation and enthalpy for the reverse step? Answer: B → A has ΔH = +25 kJ, because reversal changes the sign.

Exam focus

Write each manipulated equation and its new ΔH on the same line. Cancel species only when formula and physical state match. Sum enthalpies only after the net equation exactly matches the target, including coefficient scaling.

Advanced insight

Hess-law equation combinations are a linear-algebra problem: the target stoichiometric vector is a sum of scaled supplied vectors, and reaction enthalpy obeys the same coefficients. This viewpoint helps with larger cycles, where trial-and-error cancellation becomes cumbersome.

Summary

Reverse a thermochemical equation and reverse its ΔH sign; multiply the equation and ΔH by the same factor; add equations and their enthalpies. A valid answer requires the resulting formulae, coefficients and physical states to reproduce the target reaction exactly.

Practice questions

1. If 2A → 2B has ΔH = +60 kJ, what is ΔH for A → B? Answer: Halve the equation and enthalpy, giving +30 kJ. 2. If A → B is +30 kJ and B → C is −50 kJ, what is A → C? Answer: Add the steps and cancel B: ΔH = +30 − 50 = −20 kJ. 3. Why can H₂O(g) not simply cancel H₂O(l) in a Hess-law sum? Answer: They are different physical states with a nonzero vaporisation enthalpy between them.