Let’s try a question from Friday... When 50 mL of 1.0 mol/L hydrochloric acid is neutralized completely by 75 mL of 1.0 mol/L sodium hydroxide in a polystyrene.

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IB Topics 5 & 15 PART 1: Heat and Calorimetry
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PART 1: Heat and Calorimetry
Presentation transcript:

Let’s try a question from Friday... When 50 mL of 1.0 mol/L hydrochloric acid is neutralized completely by 75 mL of 1.0 mol/L sodium hydroxide in a polystyrene cup calorimeter, the temperature of the total solution changes from 20.2°C to 25.6°C. Determine the enthalpy change that occurs in the chemical system. Δ r H = Q = mc Δt =(50+75)g x 4.19 J/g°C x ( )°C =2.83 kJ The enthalpy change for the neutralization of HCl in this context is recorder as kJ because it’s an exothermic reaction.

Molar Enthalpies Sometimes it is better to know the enthalpy change per unit chemical amount (i.e. per mole) so that you can relate the enthalpy change to a chemical reaction equation that is balanced in moles. Molar enthalpy of reaction is the enthalpy change in a chemical system per unit chemical amount (per mole) of a specified chemical undergoing change in the system at constant pressure.

Δ r H = nΔ r H m Enthalpy change of reaction Chemical amount (moles) Molar enthalpy of reaction Note: the ‘r’s can be switched with a ‘c’s for combustion reactions

Where do I find molar quantities Molar quantities are often referenced or memorized Atomic molar masses are from the periodic table molar volume of any gasses at STP is 22.4 L/mol We can use these to predict, say, the enthalpy of combustion ( Δ c H)

Sample problem Predict the change in enthalpy due to the combustion of 10.0 g of propane used in a camp stove. 1) First you have to determing the change in enthalpy for the combustion Δ c H Δ c H = nΔ c H m To calculate Δ c H we need to know the chemical amount and molar enthalpy of combustion Δ c H m n= 10.0 g x 1/44.11g/mol (molar mass) Molar enthalpy of combustion of propane is: kJ/mol

Answer Δ c H = nΔ c H m = 10.0 g x 1/44.11 g/mol x ( kJ/mol) = kJ

Another problem Predict the enthalpy change due to the combustion of 10.0g of butane in a camp heater. The molar enthalpy of combustion to produce water vapour is kJ/mol Δ c H = nΔ c H m = 10.0 g x 1/58.14g/mol x ( kJ/mol) = kJ

Remember... Δ H = Q therefore nΔ c H m = mcΔt You can use this to determine an unknown but be careful, in calorimetry, the right hand side is all about water and the left is about the fuel.

Homework Questions 1-7 and 11 on page 494