Question

Part A) A 0.5865-g sample of lactic acid (HC3H5O3) is burned in a calorimeter whose heat capacity is 4.812 kJ/°C. The temperature increases from 23.10 °C to 24.95 °C. Calculate the heat of combustion of lactic acid per mole.

Part B) The enthalpy of reaction for the combustion of C to CO₂ is –393.5 kJ/mol C, and the

enthalpy for the combustion of CO to CO₂ is –283.0 kJ/mol.

Calculate the enthalpy for the combustion of C to CO:

Part C) Carbon occurs in two forms, graphite and diamond. The enthalpy of the combustion

of graphite is –393.5 kJ/mol and that of diamond is –395.4 kJ/mol

C (s, graphite) + O₂(g) --> CO₂ (g)   

C (s, diamond) + O₂(g) --> CO₂ (g)   

Calculate for the conversion of graphite to diamond:

∆H = ____ kJ

Answer 1

Part A)
Q = C* delta T
= 4.812 kJ/°C * (24.95-23.1)°C
=8.9022 KJ
  
molar mass of HC3H5O3= 90 g/mol
number of moles,n = mass/molar mass
= 0.5865/90
               =0.00652 mol
              
haet of combustion = -Q/n
= -8.9022 KJ/0.00652 mol
= -1366.1 KJ/mol  
Answer: -1366.1 KJ/mol                            

Part B)
delta H = delta H (C--> CO2) - delta H (CO--> CO2)
= –393.5 kJ/mol + 283.0 kJ/mol
       = -110.5 KJ/mol
Answer: -110.5 KJ/mol

Part C)
C (s, graphite) + O2(g) --> CO2 (g) ....rxn 1
C (s, diamond) + O2(g) --> CO2 (g) ...rxn 2

Reaction is C (s, graphite) ---> C (s, diamond)
It can be written as rxn 1 - rxn 2
so,
delta H = delta H1 - delta H2
= –393.5 kJ/mol + 395.4 kJ/mol
       = 1.9 KJ/mol
Answer: 1.9 KJ