Ibdp Chemistry Ia Enthalpy Change of Neutralisation

2643 Words Aug 22nd, 2013 11 Pages
Aim:

To calculate the enthalpy change of neutralization of the given pairs of acid and base.

Theory:

When alkali neutralizes an acid, a salt and water are formed. Aqueous hydrogen ions, H+(aq) from the acid react with the hydroxide ions, OH-(aq) from the alkali, forming water.

Ionic equation: H+ (aq)+OH- (aq) → H2O (l)

The identity of the salt will depend on the nature of the acid and alkali used.

The combination of H+ and OH- ions in this way releases energy. In this practical, the enthalpy changes accompanying different neutralization reactions will be measured. It is because the number of moles of water formed varies according to the acid and alkali used, it is the convention to measure enthalpy change of neutralization in kJ
…show more content…
*Uncertainty of time = 0.1s, which includes the uncertainty of the stopwatch itself (0.01s) and human reaction time. Therefore, converting uncertainty of time from s to min,
Uncertainty of time = 0.160 min
= 0.00167
≈ ±0.002 min

Analysis of Data

Graph 1: Graph of T against t when HCl is used to neutralize NaOH solution

Graph 2: Graph of T against t when CH3COOH is used to neutralize NaOH solution

A. Calculation of the Enthalpy change of neutralization

Heat released, H = mcΔθ,
Where m = mass of the solution = ρ (VHCl+VNaOH)
In this experiment, a few assumptions are made, that is

* Density of the solution, ρ= 1 g cm-3 * Specific heat capacity of the solution = 4.18 J g-1 oC-1 * The maximum temperature is reached at 3rd minute * By extrapolating the graph to t = 3min, heat lost to the surrounding and slow response time of thermometer are compensated.

Heat released during the neutralization between HCl and NaOH

From the equation in Graph 1 when t=3min, the temperature obtained is
T = (−0.3929t + 37.464) ± 0.5
T = (−0.3929 (3) + 37.464) ± 0.5
T = 36.2853 ± 0.5
T ≈ 36.3 ± 0.5 oC

Therefore, we assume that the maximum T is 36.3±0.5 oC of the reaction,

Hence,

H = ρ(VHCl+VNaOH) × c × Δθ = 1[(25.0±0.5)+(25.0±0.5)] × 4.18 × [(36.3±0.5) − (29.0±0.5)] = 1(50.0±1.0)(4.18)(7.3±1.0) = 1526 ± 1.050.0+1.07.4×100% = 1526 ± 15.51% = 1526 ± 237 ≈ 1530 ± 240 ≈ 1.53 ± 0.24 kJ