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Module 4 · L9 of 13 ~35 min ⚡ +50 XP in Learn · +25 to complete

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.

Today's hook, In 1937, Hans Krebs mapped the citric acid cycle that respiration uses to oxidise glucose. The overall reaction releases about −2803 kJ mol⁻¹ (the standard enthalpy of combustion of glucose). Written as overall net equations, photosynthesis is the reverse, so for a Hess-law comparison its enthalpy is about +2803 kJ mol⁻¹. Equal and opposite, because enthalpy is a state function.

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Where this lesson fits

Application · IQ2

Lesson question: how do photosynthesis and respiration relate energetically, and why does knowing the ΔH of one immediately give you the ΔH of the other?

1Recall Hess's Law. Enthalpy is a state function, so reversing a thermochemical equation flips the sign of ΔH while keeping its magnitude (Lesson 8).
2Recognise the reverse pair. Written as overall net equations, photosynthesis is exactly respiration run backwards, same species, opposite roles.
3Apply equal-and-opposite ΔH. Respiration ΔH ≈ −2803 kJ mol⁻¹, so photosynthesis ΔH ≈ +2803 kJ mol⁻¹, deduced, not re-measured.

Quick prerequisite: Hess's Law and equation manipulation from Lesson 8, reversing an equation flips the sign of ΔH. If that is shaky, start with the Supported route later in this lesson. Curriculum note: this lesson is an application of Hess's Law (ACSCH074), not a new dot point. Most of it is retrieval, the Hess manipulation you already met, applied to a named example; the new part is the photosynthesis/respiration context.

Know what matters most

Must know

Photosynthesis and respiration are reverse reactions, so their ΔH values are equal in magnitude and opposite in sign
Applying Hess's Law to the glucose cycle: use one ΔH (e.g. respiration −2803) to deduce the other (+2803)
The cycle closes, the two ΔH values sum to zero because enthalpy is a state function

Should know

The biological context: respiration releases stored solar energy; photosynthesis stores it (endothermic)
The ΔH values (≈ ±2803 kJ mol⁻¹) and that photosynthesis is driven by light, not heat
An alternative ΔH°_f route gives the same answer, a second Hess's-Law pathway

Going deeper

ATP coupling and Gibbs free energy (ΔG), a preview of Lesson 13, not required for this application
The multi-step biochemical pathways themselves are NOT examinable; only the overall net equations matter

Warm up and recall

Recall, your gut answer first

+5 XP warm-up

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.

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What you'll master, and the words for it

What you'll master

Know

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

Understand

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)

Can Do

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

Key terms

Photosynthesis (overall)

6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g); ΔH° ≈ +2803 kJ mol⁻¹ (endothermic; driven by light energy).

Cellular respiration (overall)

C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l); ΔH° ≈ −2803 kJ mol⁻¹ (overall net equation reverses photosynthesis).

Reverse relationship

Photosynthesis and respiration are reverse reactions; ΔH values are equal and opposite; demonstrates Hess's Law.

ATP (adenosine triphosphate)

The energy currency of cells; energy from respiration is used to synthesise ATP; ATP hydrolysis releases energy for biological work.

Multi-step pathway

Hess's Law allows breaking complex biochemical reactions into measurable steps; ΔH(overall) = sum of ΔH for each step.

Carbon cycle

CO₂ fixed in photosynthesis is released in respiration; combustion of fossil fuels disrupts the balance by adding extra CO₂.

Warm up and recall

Printable worksheet

Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.

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Cross-lesson links: This lesson applies the Hess's Law manipulation skills from L08 to a real biochemical system. The photosynthesis/respiration energy cycle demonstrates the "reverse reaction → reverse sign" rule and multi-step ΔH summation in a biological context. The ATP coupling concept here, pairing an endothermic reaction with an exothermic one to achieve a net negative ΔG, is the molecular-level version of Hess's Law that returns in L13 when Gibbs free energy explains how spontaneity is determined.

Constructing the Hess's Law Energy Cycle

Must know core concept

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

Common diagram error: Drawing the photosynthesis arrow pointing downward (as if it releases energy). Photosynthesis is strongly endothermic (ΔH = +2803 kJ mol⁻¹), glucose + O₂ sit at higher enthalpy than CO₂ + H₂O. The arrow must point upward, requiring energy input. If you draw it downward, your diagram contradicts the sign of ΔH.

Going deeper: an alternative ΔH°_f route to the same answer

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:

ΔH(respiration) = ΣΔH°_f(products) − ΣΔH°_f(reactants)
= [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?

ATP Coupling, Hess's Law in Biology

Should know core concept

Beyond the syllabus. The Hess's Law cycle in this lesson, using photosynthesis and respiration as forward and reverse routes between the same two states, is Module 4 Core and is examined. ATP coupling is not. It is here as a biological application, and no Module 4 dot-point requires you to reproduce it, so nothing you are assessed on depends on remembering the −30.5 kJ mol⁻¹ hydrolysis figure or the two-ATP arithmetic below. Carry one thing away if you read it: coupling is decided by Gibbs free energy, ΔG, not by ΔH, which is why this card says ΔG throughout and why the formal treatment waits until Lesson 13.

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.

Your body is a Hess's Law machine. Right now, your muscles are contracting (endothermic), your ribosomes are synthesising proteins (endothermic), and your ion pumps are maintaining membrane potential (endothermic). None of these violate thermodynamics, each is coupled to the favourable hydrolysis of ATP, making the combined Gibbs free energy (ΔG) negative. This is why biologists say "ATP is the energy currency of the cell."

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:

        Endothermic reaction:            A → B    ΔH = +45 kJ mol⁻¹

        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.

Why ATP stores "just the right amount" of energy: The hydrolysis of ATP releases ≈30.5 kJ mol⁻¹, small enough to be released in controlled steps without generating excess heat that would denature proteins, but large enough to drive most biochemical reactions. This is why ATP, not glucose directly, powers cellular work. Glucose releases 2803 kJ mol⁻¹ all at once, far too much for a cell to handle without burning up.

Coupling is a Gibbs-energy idea. The full criterion for spontaneity is ΔG = ΔH − TΔS (covered in Lesson 13). The coupled example above has a combined ΔG°′ = −16 kJ mol⁻¹, which is thermodynamically favourable. We use Gibbs energies (ΔG°′) here because biochemical coupling is governed by Gibbs free energy, not by enthalpy alone.

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).

Same science, choose your support

06b

Same science, choose your support

pick one

All three routes practise the same Must-Know skill: use one known ΔH to deduce the ΔH of the reverse reaction. Do one; try another if you want more practice.

Supported word bank + sentence frame start here if unsure

Word bank: reverse · +2803 · opposite · endothermic

Cellular respiration has ΔH = −2803 kJ mol⁻¹. Complete the frame:

"Photosynthesis is the ______ of respiration, so by Hess's Law its ΔH has the same magnitude but the ______ sign: ΔH = ______ kJ mol⁻¹. A positive value means photosynthesis is ______."

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Show a hint

Reversing a thermochemical equation keeps the number the same and flips only the sign: −2803 becomes +2803.

Core independent HSC-standard the exam standard

The combustion (respiration) of glucose has ΔH = −2803 kJ mol⁻¹. State ΔH for photosynthesis, justify your answer using Hess's Law, and explain why the two magnitudes must be equal.

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Stretch reasoning in a new context push yourself

A student argues: "The real biochemical pathways of respiration and photosynthesis are completely different multi-step processes, so their ΔH values cannot be exactly equal and opposite." Using Hess's Law and the idea of a state function, evaluate this claim, then state ΔH(photosynthesis) given respiration ΔH = −2803 kJ mol⁻¹.

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Worked examples

Worked example · reveal as you go

Worked example 1 +5 XP on full reveal

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 / FIND
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

Always identify what is given and what needs to be found before applying Hess's Law.

Step (a), Apply Hess's Law: reverse the equation, flip the sign
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

By Hess's Law, reversing a thermochemical equation reverses the sign of ΔH. Photosynthesis is therefore endothermic, energy must be absorbed to build glucose from CO₂ and H₂O.

Step (b), Verify the cycle closes
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

Cycle sum = 0 confirms Hess's Law is satisfied. No net chemical change; no net energy change. The system returns to its starting state.

Step (c), Biological significance
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.

The equal-and-opposite relationship is not a biological coincidence, it is a mathematical necessity arising from enthalpy being a state function.

Key idea

Predict then reveal +8 XP

1 · Predict

2 · Reveal

3 · Compare

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?

Confidence 50%

What you'll master, and the words for it

Formula reference · this lesson

core formula

📐

Formula Reference, This Lesson

Photosynthesis: $6\text{CO}_2(\text{g}) + 6\text{H}_2\text{O(l)} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6(\text{s}) + 6\text{O}_2(\text{g})$ $\Delta H = +2803 \text{ kJ mol}^{-1}$

Endothermic, requires energy input (sunlight absorbed by chlorophyll)

Respiration: $\text{C}_6\text{H}_{12}\text{O}_6(\text{s}) + 6\text{O}_2(\text{g}) \rightarrow 6\text{CO}_2(\text{g}) + 6\text{H}_2\text{O(l)}$ $\Delta H = -2803 \text{ kJ mol}^{-1}$

Exothermic; its overall net equation is the reverse of photosynthesis, so ΔH is equal and opposite by Hess's Law

ATP hydrolysis: $\text{ATP} + \text{H}_2\text{O} \rightarrow \text{ADP} + \text{P}_i$ $\Delta H \approx -30.5 \text{ kJ mol}^{-1}$ per mole

ATP coupling adds the favourable ATP hydrolysis to an unfavourable biosynthesis to make the combined Gibbs free energy (ΔG) negative

"Photosynthesis releases energy, plants produce glucose"

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.

Drill, then revisit

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.

Describe how you would draw the two arrows in a Hess's Law energy cycle diagram, direction, label, and ΔH value for each.

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.

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.

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?

Revisit your thinking

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.

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Interactive Tool, Hess's Law & Bond Energy Open fullscreen ↗

True or false?

According to the Hess’s Law tool, the total enthalpy change of a reaction is independent of the pathway taken.

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01b

Misconceptions to fix before short answer

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.

Short answer

ApplyBand 4

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

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UnderstandBand 3

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

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EvaluateBand 5

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

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Comprehensive Answers

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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.

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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