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Unit 1 PPA 3 ENTHALPY OF COMBUSTION. ENTHALPY OF COMBUSTION (Unit 1 PPA3) (1) Write the balanced equation for the enthalpy of combustion of ethanol. (2)Draw.

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Presentation on theme: "Unit 1 PPA 3 ENTHALPY OF COMBUSTION. ENTHALPY OF COMBUSTION (Unit 1 PPA3) (1) Write the balanced equation for the enthalpy of combustion of ethanol. (2)Draw."— Presentation transcript:

1 Unit 1 PPA 3 ENTHALPY OF COMBUSTION

2 ENTHALPY OF COMBUSTION (Unit 1 PPA3) (1) Write the balanced equation for the enthalpy of combustion of ethanol. (2)Draw a labelled diagram of the apparatus used in this experiment. (3) The value for the enthalpy of combustion determined using this apparatus is much less than the accepted value. Give two reasons for this lower value. 3. Heat losses to the surroundings. and Incomplete combustion of ethanol. The enthalpy of combustion is the energy released when one mole of a substance is completely burned in oxygen. The aim of this experiment is to determine the enthalpy of combustion of ethanol. A measured mass of ethanol is burned in a spirit burner and the heat released is transferred to a copper can containing a known volume of ethanol. From the resulting temperature rise, the enthalpy of combustion of ethanol can be calculated. (1) C 2 H 5 OH(l) + 3O 2 (g) 2CO 2 (g) + 3H 2 O(l) (2) Ethanol in spirit burner known mass of water thermometer

3 (5) The enthalpy change during the reaction is calculated using  H = - cm  T (a) (i) What is c ? (ii) What is the value of c for water? (b) If 200 cm 3 of water is heated, what is the value of m? (5) (a) (i) c is the specific heat capacity of water. (ii) 4.18 J kg -1 oC -1. (b) m = 0.2 kg (4) What measurements are taken during this experiment? (4) The initial water temperature. The highest water temperature reached. The mass of water being heated (in kg). The mass of ethanol burned. (6) In an experiment it was found that the temperature of 200 cm 3 of water rose from 20.1 o C to 32.5 o C when 0.42g of ethanol was burned. Calculate the enthalpy of combustion of ethanol. (6)  H = - cm  T = x 0.2 x 12.4 = kJ When 0.42 g of ethanol burns  H = kJ So when 1 mole, 46g, of ethanol burns  H = x 46 / 0.42 kJ = kJ


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