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Published byGervais Hart Modified over 9 years ago
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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 (25.6-20.2)°C =2.83 kJ The enthalpy change for the neutralization of HCl in this context is recorder as -2.83 kJ because it’s an exothermic reaction.
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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.
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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
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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)
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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: -2043.9 kJ/mol
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Answer Δ c H = nΔ c H m = 10.0 g x 1/44.11 g/mol x (-2043.9 kJ/mol) = - 463 kJ
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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 -2657.3 kJ/mol Δ c H = nΔ c H m = 10.0 g x 1/58.14g/mol x (-2657.3 kJ/mol) = - 457 kJ
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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.
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Homework Questions 1-7 and 11 on page 494
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