Calculating+enthalpy+changes

The heat energy released when a substance cools or gained when a substance increases in temperature can be calculated using the equation:

**Heat energy = m.c. Δθ ** Where //m is the mass//, //c is the specific heat capacity of the substance// and //Δθ is the temperature change of the substance//.

Usually a technique known as //**calorimetry**// is used to measure the enthalpy change associated with a particular reaction. In this technique the temperature change of a liquid inside a well insulated container is measured. In using the calorimetry technique the assumption made is that all the heat absorbed or evolved during the reaction, cause the temperature of the liquid in the calorimeter to change. (so there is no heat lost/gained from the surroundings). However, this assumption is pretty wrong since when a reaction occurs there is always heat lost or gained from the surroundings. This loss/gain of heat from the surroundings is one source of experimental error when doing experiments on enthalpy changes.

In order to reduce the uncertainties caused due to heat losses/gain form the surrounding the following technique is used. For example if the reaction is exothermic the temperature of the water in the calorimeter increases.
 * Firstly the initial temperature of the water is measured. Then as the reaction start the temperature change of the water is noted at regular time intervals. This data is used to draw a graph of temperature versus time.



//T 0 is the initial temperature of the water. As the reaction proceeds the temperature increases along the curve and reaches T 1. After that the water temperature starts to decrease along the line. In order to make up for heat losses to the surroundings, rather than taking T 1 as the final temperature, the straight cooling line is extended backwards until the time at which the reaction started, reaching T 2 (this shows the temperature which would have been reached if the reaction was instantaneous). Thus the temperature change Δθ is the difference between T 2 and T 0. //

Usually chemical reactions occur in aqueous solutions and the energy evolved alters the temperature of the water and the reactants dissolved in it. However when calculating the amount of energy gained, the energy gained by the water only is calculated, since water has a high specific heat capacity, the heat gained by the substances (reactants) dissolved in the water is ignored.

Eg.

20 cm 3 of 2 mol dm -3 NaOH is added to 30 cm 3 of 2 mol dm -3 HCl. The temperature of the mixture increases by 12 o C. Calculat the enthalpy change of the reaction?


 * As the reaction is in aqueous solution, the total volume of water= 20 +30 = 50 cm 3
 * The mass of the solution can be calculated since we know that the density of water is 1 g cm <span style="font-family: Tahoma,sans-serif; font-size: 14pt; vertical-align: super;">-3

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> //Mass of water// = 50 x 1= 50 g


 * <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">The heat energy gained by the water can be calculated as follows:

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> //Heat energy//= mc Δθ= 50g x 4.18 J g -1 o C -1 x 12 o C = 2508J = 2.508 KJ


 * <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">We know that the amount of heat released in the reaction is the same as the amount of heat gained by the water.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Now write down the equation of the reaction and identify the limiting reagent:

//<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">The balanced equation is: //

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">NaOH + HCl ---> H 2 O + NaCl <span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Mol of NaOH = (20/1000)x2=0.04 mol

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Mol of HCl = (30/1000)x2=0.06 mol

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">NaOH is the limiting reagent so all of it will react. Therefore we are going to use the amount of NaOH to calculate the enthalpy change of the reaction.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">The reaction release 2.508 KJ of energy and we have 0.04 mol of NaOH. Therefore the amount of energy released per mol of NaOH is

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">= 2.508/0.04 = **62.7 KJ mol -1 **

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">Therefore the enthalpy change of the reaction is;

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;"> **ΔH= - 62.7 KJ mol -1 ** ( the negative sign indicates that this is an exothermic reaction)


 * <span style="color: #0000ff; font-family: Tahoma,sans-serif; font-size: 10pt;">By the end of this lesson you should be able to: **
 * **<span style="color: #0000ff; font-family: Tahoma,sans-serif; font-size: 10pt;">Calculate the energy change when the temperature of a pure substance is changed. **
 * **<span style="color: #0000ff; font-family: Tahoma,sans-serif; font-size: 10pt;">Calculate the enthalpy change of a reaction using given data about the temperature change and mass of water and the reactants. **
 * **<span style="color: #0000ff; font-family: Tahoma,sans-serif; font-size: 10pt;">Remember the assumptions made when using the calorimetry technique. **
 * **<span style="color: #0000ff; font-family: Tahoma,sans-serif;">know how the effect of heat lost to surroundings is minimized by using a temp Vs time graph to find the final temperature. **
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