# Calculating Enthalpy Changes

• Created by: chunks-42
• Created on: 17-04-15 17:02

## You can find out Enthalpy Changes in the Lab

1. To measure the enthalpy change for a reaction, you only need to to know two things: The number of moles of the stuff that's reacting, The change in temperature.

2. How you go about doing the experiment depends on what type of reaction it is. Some reactions will quite happly take place in a container and you can just stick a thermometer in to find out the temperature change. It's best to use a polystyrene beaker, so that you don't lose or gain much heat through the sides.

3. Combustion reactions are trickier because the reactant is burned in air. A copper calorimeter containing a known mass of water is often used. You burna known mass of the reactant and record the temperature change of the water.

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## Calculate Enthalpy Changes Using Equation q=mc^T

It seems there's a snazzy eqauation for everything these days, and enthalpy change is no exception.

q=mc^T

q= heatlost or gained (in joules). This is the same as the enthalpy change if the pressure is constant.

m= mass of water in the calorimeter, or solution in the polystyrene beaker (in grams).

c= specific heat capacity of water (4.18 J g-1 K-1)

^T= the change in temperature of the water or solution.

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## q=mc^T example

In a laboratory experiment, 1.16g of an organic liquid fuel was completely burned in oxygen. The heat formed during this combuston raised the temperature of 100g of water from 295.3K to 357.8K. Calculate the standard enthalpy change of combustion of the fuel. Its Mr is 58.

1. First off, you need to calculate the amount of heat given out by the fuel using q=mc^T.

q=mc^T= 100x 4.18 x (357.8-295.3) = 26125 J = 26.125 kJ

2. Next you need to find out how many moles o the fuel produced this heat. It's back to the old moles=mass/Mr equation.

n=1.16/58=0.02 moles of fuel

3. The standard enthalpy of combustion involves 1 mole of fuel. So the heat produced by 1 mole of fuel = -26.125/0.02 = -1306 kJ mol-1 ( It's negative because combustion is an exothermic reaction)

The actual enthalpy change of combustion is -1615 kJ mol-1 - loads of heat has been lost and not measured. E.g. it's like a fair bit would escape through the copper calorimeter and also the fuel might not combust properly.

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## Hess's Law

Hess's Law states that:

The total enthalpy change of a reaction is always the same, no matter which route taken.

This law is handy for working out enthalpy changes that you can't find directly by doing an experiment.

Here's an example:

2NO2 (g) --> N2 (g) + 2O2 (g)

^ +114.4 kJ          ^ -180 kJ

The total enthalpy chnage for route 1 is the same as for route 2.

So the total enthalpy change for route 1 = 144.4 + (-180)= -66.4 kJ mol-1

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## Hess's Law: enthalpy of formation

Here's how to calculate the enthalpy change of formation for this reacion:SO2 + 2H2S -> 3S + 2H20

^Hf for SO2 = -297 kJ mol-1

^Hf for H2S = -20.2 kJ mol-1

^Hf for H20 = -286 kJ mol-1

Using Hess's Law: Route 1 = Route 2

So, ^Hf = the sum of products - the sum of reactants

= [0 + (-286 x 2)] - [-297 + (-20.2 x 2)] = -234.6 kJ mol-1

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## Hess's Law: enthalpy of combustion

Here's how to calculate the enthalpy change of combustion of ethanol:

2C + 3H2 +1/2 O2---> C2H5OH

^Hc (C)= -394 kJ mol-1

^Hc (H2) = -286 kJ mol-1

^Hc (ethanol) = -1367 kJ mol-1

Using Hess's Law: Route 1= Route 2

^HF (ethanol) + ^Hc (ethanol) = 2^Hc (C) + 3^Hc (H2)

^Hf (ethanol) + (-1367)= (2x -394) + (3 x -286)

^Hf (ethanol) = -788 + -858 - (-1367)= -279 kJ mol-1

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