Originally posted by Oceane:
OK to sum it up... these are the 2 questions left that I require help:
A: Propane is a versatile energy source that has several residential, commercial and industrial uses. The combustion of propane produces a substantial amount of heat energy.
C3H8 + 5O2 --> 3CO2 + 4H2O
Calculate the enthalpy change of the reaction given that an overall of 1109.5 KJ of heat is produced and that 66g of CO2 is evolved in the combustion of propane.
B: 51.5 kJ of heat was absorbed in the dissociation of HBr to give hydrogen and bromine with the evolution of 80g of gaseous bromine.
2HBr --> H2 + Br2
Calculate the bond energy of the H-Br bond.
No full solutions but I shall give you guide to solving the question:
A -> enthalpy change of the reaction = enthalpy change per 'mole' of reaction
Therefore, 66g of CO2 = how many moles of CO2? Let this answer be A1
From A1, you calculate the no. of 'moles' of the reaction (A2). A2 moles of reaction gives off 1109.5 KJ of heat, so 1 mole of reaction gives of? (1/A2 x 1109.5 KJ)
Ans: -2219 KJ per mole (3sf)
B-> As per question A. 80g of Br2 = how many moles of Br2? (B1).
From B1, you can thus calculate the no. of 'moles' of reaction (B2).
Ok next, the net enthalpy change of the reaction = bonds formed - bonds broken. In this case, you form B2 amounts of H-H and B-B bonds when you break 2 x B2 amounts of H-Br bonds.
For your info, B.E. of H-H = 436 kj per mole, B.E. of Br-Br = 193 kj per mole.
Therefore, enthalpy change = BE[H-H] + BE[Br-Br] - 2xBE[H-Br]
Rearranging, 2xBE[H-Br] = BE[H-H] + BE[Br-Br] - enthalpy change
So in this case, 2 x B2 x BE[H-Br] = B2 x BE[H-H] + B2 x BE[Br-Br] - (+51.5 kj per mole)
Sub in the values of B2 and manipulate to get your answer.
Answer: +263 kj per mole (differs from the data booklet value of +366 kj per mole though)