Hess's Law Applied, Photosynthesis & Respiration
In 1937, Hans Krebs at the University of Sheffield mapped the citric acid cycle, the central pathway by which cellular respiration oxidises glucose. The overall reaction, glucose + oxygen → carbon dioxide + water, releases about −2803 kJ mol⁻¹, the standard enthalpy of combustion of glucose, a value that can be checked with Hess's Law from tabulated data rather than measured in one step. Written as overall net equations, photosynthesis is the reverse process, so for a Hess-law comparison its enthalpy is about +2803 kJ mol⁻¹. The magnitudes match because enthalpy is a state function, not by biological coincidence.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
Every cell in your body right now is performing cellular respiration, breaking down glucose to release energy as ATP. Plants do the reverse, using sunlight to build glucose from CO₂ and water. Both processes involve the same molecules: glucose, CO₂, H₂O, and oxygen.
If the overall net equations for respiration and photosynthesis are written as reverses of each other, what does Hess's Law predict about their ΔH values?
Before this lesson: Write down: (1) What you predict ΔH(photosynthesis) and ΔH(respiration) look like relative to each other. (2) Photosynthesis is endothermic, yet plants do it continuously without any outside energy input except sunlight. Why is this not a violation of thermodynamics? Write your thinking before the lesson explains it.
Key Facts
- ΔH(photosynthesis) = +2803 kJ mol⁻¹; ΔH(respiration) = −2803 kJ mol⁻¹
- The two values are equal and opposite because the overall net equations are written as reverses
- ATP hydrolysis has ΔG°′ ≈ −30.5 kJ mol⁻¹ per mole, used to power thermodynamically unfavourable reactions
Concepts
- Why Hess's Law requires the ΔH values to be equal and opposite (reverse reaction rule)
- How a Hess's Law energy cycle connects CO₂, H₂O, and glucose at different enthalpy levels
- How ATP coupling lets organisms run thermodynamically unfavourable reactions (a Gibbs-energy idea, developed in L13)
Skills
- Apply Hess's Law to calculate ΔH(photosynthesis) from ΔH(respiration) and vice versa
- Draw and label a Hess's Law energy cycle for the photosynthesis/respiration system
- Calculate the combined Gibbs free energy (ΔG) for an ATP-coupled reaction
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
A correctly drawn Hess's Law energy cycle for photosynthesis/respiration has two levels, two arrows, and a cycle that sums to zero, these features are non-negotiable in HSC answers.
How to draw the cycle:
- Place CO₂(g) + H₂O(l) at the lower enthalpy level, these are the lower-energy molecules (stable combustion products)
- Place C₆H₁₂O₆(s) + O₂(g) at the upper enthalpy level, glucose has stored solar energy and sits 2803 kJ mol⁻¹ above
- Draw the photosynthesis arrow pointing upward (lower → upper; endothermic; ΔH = +2803 kJ mol⁻¹)
- Draw the respiration arrow pointing downward (upper → lower; exothermic; ΔH = −2803 kJ mol⁻¹)
- Verify: the two ΔH values sum to zero, the cycle is closed
Alternative Hess's Law cycle, via ΔH°f values:
You can also construct an energy cycle where the elements (C, H, O in their standard states) form the intermediate level. By Hess's Law:
= [6(−393.5) + 6(−285.8)] − [ΔH°f(glucose) + 6(0)]
= [−2361 + (−1714.8)] − ΔH°f(glucose)
= −4075.8 − ΔH°f(glucose)
This gives ΔH°f(glucose) = −4075.8 − (−2803) = −1272.8 kJ mol⁻¹ the enthalpy of formation of glucose, consistent with published data.
A Hess’s Law energy cycle shows two routes between the same reactants and products; ΔH must be the same by both routes. Required features: two energy levels, arrows with correct direction and sign, and ΔH values that sum to zero around the complete cycle.
Pause, copy the highlighted definition into your book before moving on.
Quick check: In the Hess's Law energy cycle for photosynthesis and respiration, which set of substances sits at the higher enthalpy level?
We just saw how energy cycles are drawn with correct arrows and ΔH relationships. That raises a question: does Hess’s Law apply in living systems, not just test-tube reactions? This card answers it → ATP coupling lets thermodynamically unfavourable biological reactions proceed by pairing them with a favourable partner (a Gibbs-energy idea, previewed here and developed in L13).
Living organisms run many thermodynamically unfavourable reactions, protein synthesis, ion pumping, muscle contraction, by coupling them to the highly favourable hydrolysis of ATP. Coupling is governed by Gibbs free energy (ΔG), introduced formally in L13: the combined ΔG of the coupled reactions is negative.
The ATP hydrolysis reaction:
ATP(aq) + H₂O(l) → ADP(aq) + Pᵢ(aq) ΔG°′ ≈ −30.5 kJ mol⁻¹ (standard Gibbs free energy of hydrolysis)
How ATP coupling works (Hess's Law logic):
Suppose a biosynthesis reaction has ΔG°′ = +45 kJ mol⁻¹ (thermodynamically unfavourable on its own). The cell couples it to the hydrolysis of 2 moles of ATP:
2 × ATP hydrolysis: 2ATP + 2H₂O → 2ADP + 2Pᵢ ΔG°′ = 2(−30.5) = −61 kJ mol⁻¹
Combined (sum of Gibbs energies): A + 2ATP + 2H₂O → B + 2ADP + 2Pᵢ ΔG°′ = +45 + (−61) = −16 kJ mol⁻¹ (favourable)
The combined reaction is exothermic overall thermodynamically favourable from an enthalpy perspective. By adding the two thermochemical equations (exactly as in Hess's Law), the cell achieves a net negative ΔH.
Organisms couple endothermic reactions (protein synthesis, ion pumping) to the highly exothermic hydrolysis of ATP (ΔH ≈ −30 kJ mol⁻¹), this is Hess’s Law at the molecular level. The summed ΔH of coupled reactions determines spontaneity even when individual steps are endothermic.
Add the highlighted point to your notes before the check below.
Explain it: A biosynthesis reaction has ΔG°′ = +55 kJ mol⁻¹. Explain in 2–3 sentences how many moles of ATP (ΔG°′ = −30.5 kJ mol⁻¹ per mol) must be coupled to make the overall reaction favourable (combined ΔG < 0).
Worked example · reveal as you go
Photosynthesis/Respiration Cycle Verification. Given that ΔH for cellular respiration = −2803 kJ mol⁻¹, (a) calculate ΔH for photosynthesis using Hess's Law; (b) verify that the Hess's Law energy cycle closes (sums to zero); (c) explain the biological significance of the equal and opposite values.
GIVEN: ΔH(respiration) = −2803 kJ mol⁻¹; respiration = C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l)
FIND: (a) ΔH(photosynthesis) | (b) Cycle verification | (c) Biological significance
The overall net equation for photosynthesis is the reverse of respiration:
Reverse the equation: 6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g)
Flip the sign of ΔH: ΔH = −(−2803) = +2803 kJ mol⁻¹
Add the two equations:
Photosynthesis: 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂ ΔH = +2803
Respiration: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O ΔH = −2803
All species cancel (each appears on both sides). ΔH total = +2803 + (−2803) = 0 kJ mol⁻¹
Plants absorb exactly 2803 kJ mol⁻¹ of solar energy to produce one mole of glucose. Animals (and plants at night) release exactly 2803 kJ mol⁻¹ when respiring one mole of glucose. The sun's energy is: absorbed by chlorophyll → stored in glucose bonds → transferred through the food chain → released as ATP in respiration. The Hess's Law relationship ensures the amount of solar energy captured equals the amount of chemical energy stored, the glucose molecule is a precise solar energy storage unit.
Aerobic respiration releases about 2803 kJ mol⁻¹ of energy. As overall net equations, photosynthesis is the reverse of respiration. Predict: will ΔH for photosynthesis be about +2803 kJ mol⁻¹, or a different value?
How close was your prediction?
Exactly right, the reverse reaction rule is core to Hess's Law.
Key: reversing an equation flips the sign but keeps the same kJ value. This is the central Hess's Law manipulation.
Formula Reference, This Lesson
Common errors · the 3 traps that cost marks
"Photosynthesis releases energy, plants produce glucose"
Students assume that because plants produce a product (glucose), the reaction must be exothermic and release energy.
Fix: Photosynthesis is strongly endothermic (ΔH = +2803 kJ mol⁻¹). Plants absorb solar energy and store it in glucose bonds, energy is not released, it is stored. In the energy cycle, photosynthesis arrow goes upward (to higher enthalpy), not downward.
"ATP coupling violates conservation of energy"
Students think that coupling an exothermic reaction to an endothermic one somehow "creates" energy to power the endothermic reaction.
Fix: No energy is created. The Gibbs energies of the two coupled reactions sum, and the total energy of the system plus surroundings is conserved. The favourable ATP hydrolysis drives the unfavourable biosynthesis. Combined ΔG = sum of both Gibbs energies, energy is redistributed, not created.
"Plants don't need energy for photosynthesis because sunlight is free"
Students write that photosynthesis doesn't require an energy input because sunlight is not a chemical reactant.
Fix: Photosynthesis requires exactly 2803 kJ mol⁻¹ of energy input per mole of glucose produced, this energy comes from sunlight absorbed by chlorophyll. Plants are open systems that continuously receive energy from the sun. The endothermic ΔH must be supplied; it just happens to come from electromagnetic radiation rather than another chemical reaction.
Quick-fire practice · 5 reps +2 XP per reveal
In the Hess's Law energy cycle for photosynthesis and respiration, which set of substances sits at the higher enthalpy level, and which sits at the lower level? Justify your answer using the ΔH value for photosynthesis.
Lower enthalpy level: 6CO₂(g) + 6H₂O(l), the combustion products sit at lower enthalpy.
Justification: ΔH(photosynthesis) = +2803 kJ mol⁻¹ (endothermic). Photosynthesis converts CO₂ + H₂O into glucose + O₂, and requires energy input, meaning the products (glucose + O₂) are 2803 kJ mol⁻¹ higher in enthalpy than the reactants (CO₂ + H₂O). "Arrow up" = endothermic = product higher than reactant.
Describe how you would draw the two arrows in a Hess's Law energy cycle diagram, direction, label, and ΔH value for each.
Respiration arrow: Downward, from the upper level (glucose + O₂) to the lower level (CO₂ + H₂O). Label: "Respiration, ΔH = −2803 kJ mol⁻¹". Arrow points down because the reaction is exothermic (products at lower enthalpy).
The two arrows form a closed loop. The sum of ΔH values around the loop = +2803 + (−2803) = 0, verifying Hess's Law.
Construct an alternative Hess's Law cycle that uses ΔH°f values to confirm ΔH(respiration) = −2803 kJ mol⁻¹. Use: ΔH°f[CO₂(g)] = −393.5; ΔH°f[H₂O(l)] = −285.8; ΔH°f[C₆H₁₂O₆(s)] = −1272.8; ΔH°f[O₂(g)] = 0 kJ mol⁻¹.
ΣΔH°f(products) = 6(−393.5) + 6(−285.8) = −2361.0 + (−1714.8) = −4075.8 kJ mol⁻¹
ΣΔH°f(reactants) = 1(−1272.8) + 6(0) = −1272.8 kJ mol⁻¹
ΔH°rxn = −4075.8 − (−1272.8) = −4075.8 + 1272.8 = −2803.0 kJ mol⁻¹ ✓
Confirmed: the ΔHf° method gives the same result as the direct measurement. This is Hess's Law, the pathway via elements (ΔHf° route) gives the same ΔH as the direct pathway.
The synthesis of alanine (an amino acid) has ΔG°′ = +42 kJ mol⁻¹ (Gibbs free energy, introduced in L13). The cell couples it to the hydrolysis of 2 mol of ATP (ΔG°′ ≈ −30.5 kJ mol⁻¹ per mol). By summing Gibbs energies, find the overall ΔG°′ of the coupled reaction.
(1) Biosynthesis of alanine: precursors → alanine ΔG°′ = +42 kJ mol⁻¹
(2) 2 × ATP hydrolysis: 2ATP + 2H₂O → 2ADP + 2Pᵢ ΔG°′ = 2(−30.5) = −61 kJ mol⁻¹
Combined: precursors + 2ATP + 2H₂O → alanine + 2ADP + 2Pᵢ
ΔG°′(combined) = +42 + (−61) = −19 kJ mol⁻¹
A negative combined ΔG means the coupled reaction is thermodynamically favourable, so the cell can "power" the unfavourable synthesis. Coupling is governed by Gibbs free energy, not enthalpy, you treat ΔG formally in L13.
Cellular respiration of 1 mol glucose makes ATP, with each mole of ATP synthesised storing about ΔG°′ ≈ +30.5 kJ mol⁻¹ of Gibbs free energy. Modern estimates put the yield at roughly 30–32 mol ATP per glucose, though older texts often quote up to 38. The complete oxidation of glucose makes about ΔG°′ ≈ −2880 kJ mol⁻¹ of Gibbs energy available.
(i) Estimate the Gibbs energy conserved in the ATP made (use about 32 mol ATP).
(ii) Estimate the fraction of the available Gibbs energy that is conserved (%).
(iii) Where does the rest of the energy go?
(ii) Fraction conserved ≈ 976 ÷ 2880 × 100% ≈ 34% (around 30–40% depending on the assumed ATP yield).
(iii) The rest is released to the surroundings, largely as heat, which helps maintain body temperature for enzyme function and is not "wasted" biologically. Note this compares Gibbs energies (not enthalpies), and the ATP yield is approximate and context-dependent.
Go back to your Think First response. Now you can evaluate precisely. The overall oxidation of glucose releases about −2803 kJ mol⁻¹ (its standard enthalpy of combustion):
- On the ΔH relationship: ΔH(photosynthesis) = +2803; ΔH(respiration) = −2803. The two overall net equations are reverses of each other, so by Hess's Law they have equal magnitude and opposite sign. This is a mathematical necessity of enthalpy being a state function, not a biological coincidence. (The actual biochemical pathways are not literal step-by-step reverses, only the overall net equations are.)
- On how plants run an endothermic reaction: Plants are open systems continuously receiving solar energy. The +2803 kJ mol⁻¹ comes from sunlight absorbed by chlorophyll, not from breaking thermodynamic laws, but from a continuous external energy input. This is why photosynthesis requires an external light-energy input overall.
- On ATP coupling: Cells apply Hess's Law at the molecular level, adding an exothermic ATP hydrolysis (ΔH = −30.5 kJ mol⁻¹) to an endothermic biosynthesis step to produce a combined negative ΔH. This is the same algebraic addition of thermochemical equations you practised in L08.
Pick your answer, then rate your confidencethat tells the system what to drill next.
Wrong: "Photosynthesis releases energy, the plant is making glucose, so energy must be released."
Right: Photosynthesis is strongly endothermic (ΔH = +2803 kJ mol⁻¹). The plant absorbs solar energy and stores it in the glucose molecule. The photosynthesis arrow in the energy cycle points upward, to higher enthalpy.
Wrong: "The ΔH values being equal and opposite is a biological coincidence."
Right: It is a mathematical necessity of Hess's Law. Enthalpy is a state function, reversing a reaction must reverse the sign exactly. The magnitudes are necessarily equal because the same bonds are broken and formed in opposite directions.
Wrong: "ATP coupling creates extra energy from nowhere."
Right: No energy is created. ATP hydrolysis is thermodynamically favourable (ΔG°′ = −30.5 kJ mol⁻¹); adding this to the unfavourable reaction gives a combined ΔG that is less positive or negative. Sum the Gibbs energies of the coupled steps.
Q6. (a) Write balanced thermochemical equations for both photosynthesis and cellular respiration. Include state symbols and ΔH values. (2 marks)
(b) Explain, using Hess's Law, why the ΔH values for these two reactions are equal in magnitude and opposite in sign. (2 marks)
4 MARKS
Q7. Photosynthesis is strongly endothermic (ΔH = +2803 kJ mol⁻¹), yet plants carry it out continuously.
(a) Explain why this does not violate the law of conservation of energy. (2 marks)
(b) Contrast this with an organism running an endothermic biochemical reaction using ATP coupling. In what way do both situations apply the same thermodynamic principle? (2 marks)
4 MARKS
Q8. A student states: "Because photosynthesis and respiration have equal and opposite ΔH values, the energy released by respiration in animals exactly equals the energy absorbed during photosynthesis in plants, so global energy is perfectly balanced."
(a) Is the student's statement about ΔH values chemically correct? Justify using Hess's Law. (2 marks)
(b) Evaluate the student's broader claim about global energy balance. What additional factors would need to be considered for this claim to be valid? (3 marks)
5 MARKS
Show comprehensive answers ▼
Drill 1–3, The Hess's Law Energy Cycle
Drill 1: Higher enthalpy level: C₆H₁₂O₆(s) + 6O₂(g). Lower enthalpy level: 6CO₂(g) + 6H₂O(l). Justification: ΔH(photosynthesis) = +2803 kJ mol⁻¹, positive = endothermic = products (glucose + O₂) at higher enthalpy than reactants (CO₂ + H₂O).
Drill 2: Photosynthesis arrow: upward from lower (CO₂ + H₂O) to upper (glucose + O₂) level; ΔH = +2803 kJ mol⁻¹. Respiration arrow: downward from upper to lower level; ΔH = −2803 kJ mol⁻¹. Cycle sum = 0, Hess's Law verified.
Drill 3: ΣΔH°f(products) = 6(−393.5) + 6(−285.8) = −4075.8 kJ mol⁻¹; ΣΔH°f(reactants) = 1(−1272.8) + 6(0) = −1272.8 kJ mol⁻¹; ΔH = −4075.8 − (−1272.8) = −2803.0 kJ mol⁻¹ ✓.
Drill 4–5, ATP Coupling
Drill 4: Add: biosynthesis (ΔH = +42) + 2×ATP hydrolysis (ΔH = −61). Combined: ΔH = +42 + (−61) = −19 kJ mol⁻¹. Enthalpy-favourable, the cell has powered the endothermic synthesis via Hess's Law.
Drill 5(i): ≈ 32 × 30.5 = ≈ 976 kJ mol⁻¹ of Gibbs energy conserved in ATP (≈ 1159 kJ mol⁻¹ if 38 ATP assumed). (ii) ≈ 976 ÷ 2880 × 100 ≈ 34% (around 30–40% depending on the assumed ATP yield). (iii) The rest is released to the surroundings, largely as heat, maintaining body temperature and supporting enzyme function. This compares Gibbs energies, and the ATP yield is approximate.
Multiple Choice
1. B Respiration is the reverse of photosynthesis. Hess's Law: reversing a reaction changes the sign of ΔH. Therefore ΔH(respiration) = −(+2803) = −2803 kJ mol⁻¹.
2. B Photosynthesis is endothermic (+2803), meaning glucose + O₂ sit at higher enthalpy than CO₂ + H₂O. Option A has the levels reversed.
3. C1 mol: +55 − 30.5 = +24.5 (still positive). 2 mol: +55 − 61 = −6 kJ mol⁻¹ (negative ✓). Minimum = 2 mol.
4. A This is a direct consequence of Hess's Law and enthalpy being a state function. Mass conservation (option B) is always true but does not by itself explain the ΔH relationship.
5. DΣΔH°f(products) = 6(−393.5) + 6(−285.8) = −4075.8; ΣΔH°f(reactants) = 1(−1272.8) + 6(0) = −1272.8; ΔH = −4075.8 − (−1272.8) = −2803.0 kJ mol⁻¹. Option C is just the products sum, not the final answer.
Short Answer Model Answers
Q6 (4 marks):
(a) Photosynthesis: 6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g) ΔH = +2803 kJ mol⁻¹ [1]; Respiration: C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l) ΔH = −2803 kJ mol⁻¹ [1].
(b) The overall net equation for respiration is the reverse of the photosynthesis equation, the same reactants and products in opposite roles (the actual biochemical pathways differ) [½]. By Hess's Law, reversing a thermochemical equation multiplies ΔH by −1 [½]. Therefore ΔH(respiration) = −ΔH(photosynthesis), making the values equal in magnitude and opposite in sign [1].
Q7 (4 marks):
(a) Plants are open systems that continuously absorb energy from sunlight [1]. The 2803 kJ mol⁻¹ required for photosynthesis is supplied by solar radiation absorbed by chlorophyll, energy is not created, it is converted from electromagnetic (light) energy to chemical energy stored in glucose bonds [1]. Conservation of energy is maintained.
(b) In both cases, an unfavourable process is made thermodynamically feasible by coupling it to a favourable energy source [1]. Plants couple photosynthesis to sunlight; animals couple biosynthesis to ATP hydrolysis. In both cases the principle is the same: coupling makes the combined Gibbs free energy (ΔG) negative (or at least more negative than the unfavourable reaction alone) [1].
Q8 (5 marks):
(a) Yes, chemically correct per mole of glucose [½]. Respiration is the reverse of photosynthesis. By Hess's Law (reversing a reaction reverses ΔH), ΔH(respiration) = −ΔH(photosynthesis) = −(+2803) = −2803 kJ mol⁻¹ per mole of glucose [1]. The magnitudes are equal and the signs are opposite [½].
(b) The broader claim is oversimplified, at least three factors are missing: (i) Not all photosynthesised glucose is immediately respired, biomass accumulates (wood, fossil fuels), storing energy on geological timescales [1]; (ii) Combustion of fossil fuels releases carbon fixed by ancient photosynthesis that was not respired, adding CO₂ to the atmosphere that was previously sequestered [1]; (iii) The rates of photosynthesis and respiration globally are not equal, which is why atmospheric CO₂ levels have been changing, true global balance would require equal rates over all timescales and all organisms [1].
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