Hey all. Having a hard time figuring out how to tackle this question, help would be much appreciated!
6.652kJ of heat is required to raise the temperature of a calorimeter and its contents by 1.000 degrees Celcius. When 1.456g of butane gas is burned by the calorimeter, the temperature was raised by 10.00 degrees Celcius. Calculate the enthalpy of combustion for butane and add it to a balanced equation.
6.652kJ of heat is required to raise the temperature of a calorimeter and its contents by 1.000 degrees Celcius. When 1.456g of butane gas is burned by the calorimeter, the temperature was raised by 10.00 degrees Celcius. Calculate the enthalpy of combustion for butane and add it to a balanced equation.
-
If 6.652 kJ is required to heat it by 1 degree celsius, and 1.456g of butane combusting raised it by 10, then the energy put out by the butane must be 10*6.652kJ = 66.52kJ
butane = 58.12 g mol−1
1.456/58.12 = 0.02505 mols
so 0.02505 mols puts out 66.52kJ of heat when combusted. Therefore 1 mol would emit 2655.489kJ.
therefore the molar enthalpy would be 2660kJ/mol (3 significant figures)
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O + 5320kJ (there were two mols of C4H10, therefore there'll be twice as much energy as the molar enthalpy of 2660kJ)
butane = 58.12 g mol−1
1.456/58.12 = 0.02505 mols
so 0.02505 mols puts out 66.52kJ of heat when combusted. Therefore 1 mol would emit 2655.489kJ.
therefore the molar enthalpy would be 2660kJ/mol (3 significant figures)
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O + 5320kJ (there were two mols of C4H10, therefore there'll be twice as much energy as the molar enthalpy of 2660kJ)