Reactivity 1.1.4

Standard Enthalpy & Calorimetry

Calculating enthalpy changes from experimental data using $q=mc\Delta T$ .

To compare enthalpy changes fairly, we measure them under Standard Conditions. This is denoted by the Plimsoll line symbol ( $\theta$ or $\circ$ ).

Pressure

100 kPa

Concentration

1.0 mol dm⁻³

State

Standard State

e.g., $H_2(g), H_2O(l)$

The Calorimetry Equation

q = mc\Delta T

q

= Heat energy (Joules, J)

m

= Mass of the Surroundings (g)

Usually the water/solution getting hot.

c

= Specific Heat Capacity

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

\Delta T

= Temperature Change

$T_{final} - T_{initial}$

Deep Think Concept

Calculated $\mathbf{q}$ is just the heat absorbed by the water. To find the Molar Enthalpy Change of the reaction:

\Delta H = \frac{-q}{n}

The Negative Sign: Essential. Since $q$ is heat gained by water, $-q$ is heat lost by reaction.
n (moles): Moles of the limiting reactant (or fuel burned).
Units: $q$ is in Joules. $\Delta H$ is usually in $\text{kJ mol}^{-1}$ . Remember to divide by 1000!

Paper 2 Style

0.50 g of Methanol ( $CH_3OH$ , $M_r = 32.04$ ) is burned to heat 100 g of water.

The temperature of the water rises by 25.0 °C. Calculate the Enthalpy of Combustion ( $\Delta H_c$ ) in $kJ \: mol^{-1}$ .

Assume $c = 4.18 \: J g^{-1} K^{-1}$ for water.

[2 Marks]

The theoretical value for the enthalpy of combustion of methanol is $-726 \: kJ \: mol^{-1}$ .

Suggest two reasons why the experimental value calculated above is less exothermic (smaller magnitude) than the theoretical value.