Calculate q when 200\ g of water rises by 12.0\ ^C. [1 mark]

OCR OCR 2019-style practice: A student burned 0.250\ g of ethanol and used the heat to raise the temperature of 150\ g of water by 18.0\ ^C. Calculate the enthalpy of combustion of ethanol in kJ mol^-1. (c = 4.18\ J g^-1K^-1, M_r ethanol = 46.0.)

Heat absorbed by water q = mc T = 150 × 4.18 × 18.0 = 11,286\ J = 11.286\ kJ (1). Moles of ethanol = 0.25046.0 = 5.43 × 10^-3\ mol (1). _c H = -11.2865.43 × 10^-3 = -2078\ kJ mol^-1, about -2080\ kJ mol^-1 (1). The value is negative because combustion is exothermic (1). Markers reward q = mc T, the moles, the division to get per mole, and the negative sign.

OCR OCR 2021-style practice: Use the average bond enthalpies H-H = +436, Cl-Cl = +243 and H-Cl = +432\ kJ mol^-1 to calculate the enthalpy change for H_2 + Cl_2 → 2HCl.

Bonds broken (reactants): 436 + 243 = +679\ kJ mol^-1 (1). Bonds made (products): 2 × 432 = 864\ kJ mol^-1, released so -864\ kJ mol^-1 (1). H = (bonds broken) - (bonds made) = 679 - 864 = -185\ kJ mol^-1 (1). Markers reward bonds broken, bonds made, and the subtraction giving an exothermic value.

EnglandChemistrySyllabus dot point

How do we measure and calculate the energy released or absorbed in a chemical reaction?

Enthalpy and standard enthalpy changes, exothermic and endothermic reactions, calorimetry and the q = mcDeltaT equation, average bond enthalpies, and Hess's law including formation and combustion cycles.

An OCR H432 module 3 answer on enthalpy changes: standard enthalpy definitions, calorimetry with q = mcDeltaT, bond enthalpy calculations, and Hess's law cycles for formation and combustion.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
Standard enthalpy changes
Calorimetry
Average bond enthalpies
Hess's law
Examples in context
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What this topic is asking

OCR specification point 3.2.1 wants you to define standard enthalpy changes, classify reactions as exothermic or endothermic, measure enthalpy changes by calorimetry using $q = mc\Delta T$ , calculate enthalpy changes from average bond enthalpies, and apply Hess's law with enthalpy of formation and combustion cycles. This is the start of the physical chemistry that runs through Modules 3 and 5.

Standard enthalpy changes

In an exothermic reaction the products have lower enthalpy than the reactants, so $\Delta H$ is negative; in an endothermic reaction $\Delta H$ is positive. Energy is needed to break bonds (endothermic) and released when bonds form (exothermic).

Calorimetry

To get $\Delta H$ per mole, calculate $q$ , convert to kJ, divide by the moles reacting, and attach the correct sign (negative for a temperature rise, which signals an exothermic reaction).

Average bond enthalpies

Hess's law

Hess cycle using enthalpies of formation

Find $\Delta H$ for a reaction from formation data using $\Delta H = \sum \Delta_f H(\text{products}) - \sum \Delta_f H(\text{reactants})$ .

step 1: Write the formation cycle

Elements sit at the bottom; arrows go up to reactants and up to products. By Hess's law, going reactants to products directly equals going down to elements then up to products.

step 2: Apply the formula

For $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ with $\Delta_f H$ values $-1207$ , $-635$ and $-394\ \text{kJ mol}^{-1}$ :

\Delta H = [(-635) + (-394)] - (-1207) = -1029 + 1207 = +178\ \text{kJ mol}^{-1}.

step 3: Interpret

The positive value shows the thermal decomposition of calcium carbonate is endothermic, which is why a kiln must supply heat continuously.

Examples in context

Example 1. Comparing fuels. Calorimetry of different alcohols gives enthalpies of combustion that, per gram, let engineers compare energy density, though heat lost to surroundings always makes the measured value less exothermic than the data-book figure.

Example 2. Why bond-enthalpy answers differ. A value from average bond enthalpies rarely matches the data-book $\Delta H$ exactly, because real bonds in a specific molecule differ from the averaged values, a point OCR likes you to explain.

Try this

Q1. State what is meant by the standard enthalpy of formation of a compound. [2 marks]

Cue. The enthalpy change when one mole of the compound is formed from its constituent elements in their standard states under standard conditions.

Q2. Calculate $q$ when $200\ \text{g}$ of water rises by $12.0\ ^{\circ}\text{C}$ . [1 mark]

Cue. $q = 200 \times 4.18 \times 12.0 = 10{,}032\ \text{J} = 10.0\ \text{kJ}$ .

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20194 marksA student burned

0.250\ \text{g}

of ethanol and used the heat to raise the temperature of

150\ \text{g}

of water by

18.0\ ^{\circ}\text{C}

. Calculate the enthalpy of combustion of ethanol in

\text{kJ mol}^{-1}

. (

c = 4.18\ \text{J g}^{-1}\text{K}^{-1}

M_r

ethanol

= 46.0

Show worked answer →

Heat absorbed by water $q = mc\Delta T = 150 \times 4.18 \times 18.0 = 11{,}286\ \text{J} = 11.286\ \text{kJ}$ (1).

Moles of ethanol $= \dfrac{0.250}{46.0} = 5.43 \times 10^{-3}\ \text{mol}$ (1).

$\Delta_c H = -\dfrac{11.286}{5.43 \times 10^{-3}} = -2078\ \text{kJ mol}^{-1}$ , about $-2080\ \text{kJ mol}^{-1}$ (1). The value is negative because combustion is exothermic (1).

Markers reward $q = mc\Delta T$ , the moles, the division to get per mole, and the negative sign.

OCR 20213 marksUse the average bond enthalpies

\text{H-H} = +436

\text{Cl-Cl} = +243

and

\text{H-Cl} = +432\ \text{kJ mol}^{-1}

to calculate the enthalpy change for

\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}

Show worked answer →

Bonds broken (reactants): $436 + 243 = +679\ \text{kJ mol}^{-1}$ (1).

Bonds made (products): $2 \times 432 = 864\ \text{kJ mol}^{-1}$ , released so $-864\ \text{kJ mol}^{-1}$ (1).

$\Delta H = \text{(bonds broken)} - \text{(bonds made)} = 679 - 864 = -185\ \text{kJ mol}^{-1}$ (1).

Markers reward bonds broken, bonds made, and the subtraction giving an exothermic value.

Related dot points

Sources & how we know this

OCR A-Level Chemistry A (H432) specification — OCR (2015)