8.5-8.6 - Calorimetry

Temperature Changes And Enthalpy Signs

A thermometer does not measure enthalpy directly. It measures how the surroundings, usually a solution or a known mass of water, change temperature while a chemical system reacts.

In calorimetry the model is:

chemical reaction releases or absorbs energy → surroundings change temperature → temperature change is converted into energy transfer → energy transfer is scaled to one mole of reaction or fuel.

The sign belongs to the chemical system, not to the thermometer reading.

That is why a temperature rise often produces a negative value for $\Delta H$. The solution got warmer because the reaction transferred energy out to the surroundings.

From Temperature Change To Energy

A larger mass needs more energy for the same temperature rise; the specific heat capacity tells us how much energy one gram needs for a one-kelvin rise.

Each symbol has a practical meaning:

For aqueous mixtures, a common assumption is that the solution has density $1.00 \mathrm{g\,cm^{-3}}$. This lets you estimate:

For example, mixing $25.0 \mathrm{cm^{3}}$ of one solution with $25.0 \mathrm{cm^{3}}$ of another gives an estimated mass of $50.0\,\mathrm{g}$, if the density is assumed to be $1.00 \mathrm{g\,cm^{-3}}$.

Worked example: energy transferred to a solution

$50.0 \mathrm{cm^{3}}$ of acid is mixed with $50.0 \mathrm{cm^{3}}$ of alkali in an insulated cup. The temperature rises from $20.0 {}^{\circ}\mathrm C$ to $26.7 {}^{\circ}\mathrm C$. Assume the density of the final solution is $1.00 \mathrm{g\,cm^{-3}}$ and the specific heat capacity is $4.18 \mathrm{J\,g^{-1}\,K^{-1}}$.

Find the energy transferred to the solution.

Step 1: estimate the mass of solution.

$V = 50.0 + 50.0 = 100.0\,\mathrm{cm^{3}}$, so $m = 100.0\,\mathrm{g}$

Step 2: find the temperature change.

Step 3: use $Q = mc\Delta T$.

$Q = 100.0\times 4.18\times 6.7$
$Q = 2800.6\,\mathrm{J}$
$Q = 2.80\,\mathrm{kJ}$

Chemical interpretation: the solution gained $2.80\,\mathrm{kJ}$ of energy. If the reaction caused the temperature rise, the reaction itself lost this energy to the surroundings.

Common-error contrast

From Energy To Molar Enthalpy

An enthalpy change is normally reported per mole, so the calculation does not stop at $Q$.

Use this route:

1. Calculate $Q$ in J from $mc\Delta T$.
2. Convert $Q$ from J to kJ.
3. Work out the relevant amount in mol.
4. Divide by moles.
5. Add the correct sign and units.

For many simple calorimetry calculations:

The mole value $n$ must match the enthalpy change being asked for. For neutralisation it is often the amount of water formed or the limiting acid/alkali in the ionic equation. For combustion it is the amount of fuel burned. For reaction enthalpy it is the amount of reaction as defined by the balanced equation.

Worked example: enthalpy change of neutralisation

$50.0 \mathrm{cm^{3}}$ of $1.00 \mathrm{mol\,dm^{-3}} \ce{HCl(aq)}$ is mixed with $50.0 \mathrm{cm^{3}}$ of $1.00 \mathrm{mol\,dm^{-3}} \ce{NaOH(aq)}$. The temperature rises from $20.0 {}^{\circ}\mathrm C$ to $26.7 {}^{\circ}\mathrm C$. Assume density $1.00 \mathrm{g\,cm^{-3}}$ and specific heat capacity $4.18 \mathrm{J\,g^{-1}\,K^{-1}}$.

Calculate the enthalpy change of neutralisation in $\mathrm{kJ\,mol^{-1}}$.

The unrounded energy transfer has already been found:

$Q = 2.8006\,\mathrm{kJ}$

Now calculate the amount reacting.

$\ce{H+(aq) + OH-(aq) -> H2O(l)}$

When amounts are unequal, calculate both amounts before choosing the limiting reagent. For example, $0.0300\,\mathrm{mol}$ HCl and $0.0200\,\mathrm{mol}$ NaOH form only $0.0200\,\mathrm{mol}$ water; use the total solution mass in $Q$, but this limiting water amount in $-Q/n$. For a non-1:1 equation, divide each reactant amount by its coefficient first.

The acid and alkali react in a $1:1$ mole ratio, so $0.0500\,\mathrm{mol}$ of water is formed.

$\Delta H = -2.8006 / 0.0500 = -56.012\,\mathrm{kJ\,mol^{-1}}$
$\Delta H = -56.0 \mathrm{kJ\,mol^{-1}}$

The answer needs both sign and units:

Chemical interpretation: the negative sign agrees with the temperature rise. The value is for one mole of water formed, not for the $100.0 \mathrm{cm^{3}}$ mixture used in this one experiment.

Guided practice

Insulated-Cup Calorimetry

When substances are mixed in an insulated container, the apparatus is designed to make the solution temperature change as close as possible to the energy change caused by the reaction.

[DIAGRAM: asset_name: Temperature-time extrapolation for calorimetry; asset_slug: edexcel_a_level_chemistry_l043_temperature_extrapolation; recommended_method: deterministic_chart; description: Temperature-time extrapolation for calorimetry, code-drawn on a pale 16:9 canvas.]