A2 Chemistry Question

can anyone help me with the following question?

Question 1
Chemical companies manufacture containers filled with liquid butane for use by campers. The enthalpy change of combustion of butane is -3000kJ / per mole.

a) Write an equation for the complete combustion of butane

A camper estimates that the liquid butane left in a container would give 1.2 dm³ of butane gas (ordinary temperature and pressure)

b) Calculate the mass of water at 20ºC that could be brought to the boiling point by burning this butane; using the following information

Assume that
80% of the heat from the butane is absorbed by water
The specific heat capacity of water is 4.2 J / per gram / in Kelvin
1 mol of gas occupies 24 dm³ at ordinary temperature and pressure.

c) Suggest how the camper might have estimated how much butane was left in the container.

d) When burnt in a limited supply of air, butane forms carbon and steam.

i) Construct a balanced equation for this reaction.
ii) The enthalpy change for the above reaction is -1400kJ / pre mole. Explain why the enthalpy changes for the two combustion reactions of butane are different.
iii) What additional quantitative information can be calculated from this difference?

Reply 1

ChemistBoy

urbane

C4H10(g) + 6.5O2(g) ----> 4CO2(g) + 5H2O(g)

A camper estimates that the liquid butane left in a container would give 1.2 dm³ of butane gas (ordinary temperature and pressure)

b) Calculate the mass of water at 20ºC that could be brought to the boiling point by burning this butane; using the following information

Assume that
80% of the heat from the butane is absorbed by water
The specific heat capacity of water is 4.2 J / per gram / in Kelvin
1 mol of gas occupies 24 dm³ at ordinary temperature and pressure.

1 mole occupies 24dm3 so 1.2dm3/24dm3 = 0.0125 mol of butane.

butane gives out 3000kJ/mol during complete combustion so,

0.0125 mol give 3000kJ x 0.0125 = 37.5kJ or 37,500J

80% of the heat is adsorbed by the water = 37,500 x 0.8 = 30,000J

The water needs to be raised by 80 K to boil it so we need 4.2kJ x 80 per gram of water = 336.

30,000J/336J/g = 89.29g of water

c) Suggest how the camper might have estimated how much butane was left in the container.

Weight differential, if he or she knew the weight of the canister when full, and how much butane in total there is (should be on the label of the canister), he or she could deduce the weight of the butane used and hence that remaining.

[QUOTE]
d) When burnt in a limited supply of air, butane forms carbon and steam.

i) Construct a balanced equation for this reaction.

C4H10(g) + 2.5O2(g) ----> 4C(s) + 5H2O(g)

ii) The enthalpy change for the above reaction is -1400kJ / pre mole. Explain why the enthalpy changes for the two combustion reactions of butane are different.

Bond breaking and formation, formation of C=O releases more energy than formation carboniferous deposits.

iii) What additional quantitative information can be calculated from this difference?

Enthalpy of formation of CO2 (from its component elements at STP).

Consider the reactions

C4H10(g) + 2.5O2(g) ---> 4C(s) + 5H2O(g) DHcomb = -1400kJ/mol

C4H10(g) + 6.502(g) ---> 4CO2(g) + 5H20(g) DHcomb = -3000kJ/mol

This leaves us with -1600kJ/mol for this reaction:

4C(s) + 4O2(g) ----> 4CO2(g)

dividing by 4:

C(s) + O2(g) ----> CO2(g) DHform = -400kJ/mol

Reply 2

urbane

[QUOTE="ChemistBoy"]C4H10(g) + 6.5O2(g) ----> 4CO2(g) + 5H2O(g)

1 mole occupies 24dm3 so 1.2dm3/24dm3 = 0.0125 mol of butane.

butane gives out 3000kJ/mol during complete combustion so,

0.0125 mol give 3000kJ x 0.0125 = 37.5kJ or 37,500J

80% of the heat is adsorbed by the water = 37,500 x 0.8 = 30,000J

The water needs to be raised by 80 K to boil it so we need 4.2kJ x 80 per gram of water = 336.

30,000J/336J/g = 89.29g of water

Weight differential, if he or she knew the weight of the canister when full, and how much butane in total there is (should be on the label of the canister), he or she could deduce the weight of the butane used and hence that remaining.

d) When burnt in a limited supply of air, butane forms carbon and steam.

i) Construct a balanced equation for this reaction.

C4H10(g) + 2.5O2(g) ----> 4C(s) + 5H2O(g)

Bond breaking and formation, formation of C=O releases more energy than formation carboniferous deposits.

Enthalpy of formation of CO2 (from its component elements at STP).

Consider the reactions

C4H10(g) + 2.5O2(g) ---> 4C(s) + 5H2O(g) DHcomb = -1400kJ/mol

C4H10(g) + 6.502(g) ---> 4CO2(g) + 5H20(g) DHcomb = -3000kJ/mol

This leaves us with -1600kJ/mol for this reaction:

4C(s) + 4O2(g) ----> 4CO2(g)

dividing by 4:

C(s) + O2(g) ----> CO2(g) DHform = -400kJ/mol