Thermometer: It is part of surroundings so can measure temp change in surroundings.
• Kelvin Scale: Scale starts at absolute zero 0 K or -273oC. 1 K rise = 1
oC rise. oC + 273 = K.
• Calculating Energy Change: Energy change of surroundings is calculated using q = mc∆t. In joules.
• Mass of Surroundings: Have to identify the materials changing temperature. In grams.
• Specific Heat Capacity of ‘Surrounding’: c. The energy required to raise the temperature of 1g of
substance by 1k. Different substance require different quantities of energy to produce same
temperature change.
• In most experiments will measure temperature change of water or aqueous solution so c = 4.18 J
g
-1 K
-1
. Density of water 1.00 g cm-3Temperature Change of Surroundings: ∆t. Final subtract initial thermometer reading.
• Determination of Enthalpy Change of Combustion: ∆cH.
• Experimental Method: Record initial temperature of water to nearest 0.5 oC. Add methanol/ liquid
fuel to spirit burner. Weigh spirit burner with methanol. Place spirit burner under beaker. Stir
water. After some minutes extinguish flame and immediately record maximum temperature.
Reweigh spirit burner with methanol. Assume wick has not burnt.
• Calculating Experimental Enthalpy Change of Combustion:
– Work out mass of fuel and temperature change before and after.
– Calculate energy change of surroundings (water) q using 150 x 4.18 x ∆t. Formula gives answer
in J so divide by 1000 to get kJ. Given in Joules
o 150 cm3 of water = 150g as water has density of 1.00 g cm-3
.
– Calculate, in mol, amount of fuel burnt. Using mass/ Mr.
– Calculate ∆cH in kJ mol-1
. ∆H = Q (kJ)/ moles of fuel. In units kJ mol-1 which is enthalpy change
of combustion- gives one mol.
o So have found out that 1 mol of fuel transfers _kJ of energy to water as units are per mol.
– Multiply by 2, if the balancing number in equation is 2 to get 2 moles- see equation.
– Must determine sign of final answer ∆cH. As water has gained energy, reaction is exothermic
so a negative sign must be added on.
• Accuracy of Experimental ∆cH Value: The data book gives a value ∆cH for a fuel. Experimental
value will be different. Reasons why…
– Heat loss to surroundings other than water. Examples are beaker.
– Incomplete Combustion of Fuel: Carbon monoxide or carbon may have been produced
instead of carbon dioxide. Would see carbon as black soot on beaker.
– Evaporation of Fuel from Wick: Burner must be weighed immediately after extinguishing,
flame or fuel might have evaporated from wick. Spirt burners have cover to reduce this.
– Non- Standard Conditions: Data book value is standard value, which may not be conditions of
experiment- the only one where could lead to a value for ∆cH to be more exothermic than
expected.
• Solution:
– Use of draught screens to minimise errors from heat loss.
– Input of oxygen gas to minimise errors from incomplete combustion.
• Negative Values: -700 more exothermic and more negative but not bigger than -500 .
• Determination of an Enthalpy Change of Reaction: ∆rH. Between two solutions or solid + solution.
The solution itself is immediate surroundings.
• Can be determined using polystyrene cups- offer insulation against heat loss. If glass beaker used,
it would conduct some of energy.
• If greater volume of water or acid used, same amount of energy would heat greater volume.
• If concentration less then slower rate of reaction, but can convert concentration to moles to see
amount of product.
• Calculating ∆rH:
– Calculate energy change in solution q using m x 4.18 x ∆t. Then J/ 1000 = kJ.
– Calculate the mol of one of the substances reacted (the one not in excess and used up first).
Use mass/ Mr. If solution use c x v/ 1000.
– Divide kJ by mol to get ∆rH in kJ mol-1
. Remember negative sign if exothermic.
Cooling Curves: Take temperature of solution until constant.
Weigh out excess of another substance and add it to solution.
Record temperature every 30s until constant. Plot temperature (y
axis) against time (x axis).
• Extrapolate the cooling curve section of graph back to time when
zinc added (red line). Draw vertical line at time solutions were
mixed (blue line). Use these results to do same calculation above.
Should be more accurate than method above and corrects for
cooling.
• Remember to not just read of temperature on graph- work out
temperature change.
• Determination of Enthalpy Change of Neutralisation: ∆neutH. Two
solutions react rather than solution + solid.
• Calculating ∆neutH: Should be between 57 and 58 kJ mol-1
.
– Starting temperature- measure temp of both solutions and calculate average.
– Same as previous. However mass is the total mass of both solutions.
– Must calculate moles of water. May have to use molar ratio if know moles of the acid/ alkali
used to find out moles of water.
– Same as previous. If needed change balancing numbers of whole equation to form 1 mole of
H2O. Remember negative sign if exothermic.