A student measures the enthalpy change of neutralisation between hydrochloric acid and sodium hydroxide by mixing the two solutions in an insulated cup and recording the temperature. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Quantity Value Volume of 1.00 mol dm^-3 HCl(aq) used 50.0…

Answer: ((a)) The enthalpy change when one mole of water is formed by the neutralisation of an acid by an alkali, under standard conditions, using dilute aqueous solutions. ((b)) q = 2884.2 J = 2.88 kJ (3 s.f.) ((c)) n(H2O) = 0.0500 mol; Hneut = -57.7 kJ mol^-1 (3 s.f.) ((d)) Heat is lost from the mixture to the surroundings, the cup and the thermometer during the experiment, so less heat is measured than is actually released by the reaction.. Part (a): Defining the standard enthalpy change of neutralisation The standard enthalpy change of neutralisation, Hneut^, is the enthalpy change when one mole of water is formed by the reaction of an acid with an alkali, using dilute aqueo…

AS Chemistry 9701: Enthalpy of Neutralisation by Calorimetry (q = mcΔT)

Quantity	Value
Volume of $1.00\ \text{mol dm}^{-3}$ HCl(aq) used	$50.0\ \text{cm}^3$
Volume of $1.00\ \text{mol dm}^{-3}$ NaOH(aq) used	$50.0\ \text{cm}^3$
Initial temperature (both solutions, before mixing)	$19.0\,^\circ\text{C}$
Highest temperature reached after mixing	$25.9\,^\circ\text{C}$
Density of the resulting mixture	$1.00\ \text{g cm}^{-3}$
Specific heat capacity of the mixture, $c$	$4.18\ \text{J g}^{-1}\text{ K}^{-1}$

Quantity

Value

Volume of

1.00\ \text{mol dm}^{-3}

HCl(aq) used

50.0\ \text{cm}^3

Volume of

1.00\ \text{mol dm}^{-3}

NaOH(aq) used

50.0\ \text{cm}^3

Initial temperature (both solutions, before mixing)

19.0\,^\circ\text{C}

Highest temperature reached after mixing

25.9\,^\circ\text{C}

Density of the resulting mixture

1.00\ \text{g cm}^{-3}

Specific heat capacity of the mixture,

c

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

Part (a): Defining the standard enthalpy change of neutralisation

The standard enthalpy change of neutralisation, $\Delta H_{neut}^{\ominus}$ , is the enthalpy change when one mole of water is formed by the reaction of an acid with an alkali, using dilute aqueous solutions, under standard conditions.

Part (b): Heat energy released

The total mass of the mixture is the combined volume of the two solutions, since the density is $1.00\ \text{g cm}^{-3}$ : $m = 50.0 + 50.0 = 100\ \text{cm}^3 \Rightarrow 100\ \text{g}$

The temperature rise is: $\Delta T = 25.9 - 19.0 = 6.9\,^\circ\text{C}\ (= 6.9\ \text{K})$

Using $q = mc\Delta T$ : $q = 100 \times 4.18 \times 6.9$

Working in two steps: $100 \times 4.18 = 418\ \text{J K}^{-1}$ $418 \times 6.9 = 2884.2\ \text{J}$

So $q = 2884.2\ \text{J} = 2.88\ \text{kJ}$ (3 s.f.).

Check (independent recomputation): $418\times6.9 = 418\times7 - 418\times0.1 = 2926 - 41.8 = 2884.2\ \text{J}$ , consistent.

Part (c): Amount of water formed and $\Delta H_{neut}$

Amount of HCl: $n(\text{HCl}) = \frac{50.0}{1000} \times 1.00 = 0.0500\ \text{mol}$

Amount of NaOH: $n(\text{NaOH}) = \frac{50.0}{1000} \times 1.00 = 0.0500\ \text{mol}$

The reaction is exactly $1:1$ , so both reagents are used up completely and exactly $0.0500\ \text{mol}$ of water is formed.

The heat calculated in (b) was released by forming this amount of water, so: $\Delta H_{neut} = -\frac{q}{n} = -\frac{2.8842\ \text{kJ}}{0.0500\ \text{mol}}$

Dividing: $\frac{2.8842}{0.0500} = 57.684$

so $\Delta H_{neut} = -57.684\ \text{kJ mol}^{-1} \approx -57.7\ \text{kJ mol}^{-1}\ (3\ \text{s.f.})$

The value is negative because neutralisation is exothermic. Heat is released to the solution, raising its temperature.

Check (independent recomputation): $2884.2\ \text{J}\div0.0500\ \text{mol}=57\,684\ \text{J mol}^{-1}=57.684\ \text{kJ mol}^{-1}$ , confirming $\Delta H_{neut}\approx-57.7\ \text{kJ mol}^{-1}$ .

Part (d): Why the experimental value is smaller in magnitude

Not all of the heat released by the reaction is retained by the solution: some heat escapes to the surroundings through the walls of the cup, and some warms the cup itself and the thermometer rather than the solution. This means less heat is measured than the reaction actually releases, so the experimental $\Delta H_{neut}$ is less exothermic (smaller in magnitude) than the true value.

Final answers

(a) Enthalpy change when one mole of water forms from the neutralisation of an acid by an alkali, in dilute aqueous solution, under standard conditions.
(b) $q = \boxed{2884.2\ \text{J}} = 2.88\ \text{kJ}$
(c) $n(\text{H}_2\text{O}) = \boxed{0.0500\ \text{mol}}$ ; $\Delta H_{neut} = \boxed{-57.7\ \text{kJ mol}^{-1}}$ (3 s.f.)
(d) Heat losses to the surroundings and apparatus mean less energy is measured than the reaction truly releases.