What is the molar heat of combustion when #"25000 g"# of methanol is combusted in a bomb calorimeter to raise the temperature of the surrounding water by #40^@ "C"# if the heat capacity of the calorimeter is #"10.4 kJ/"^@ "C"#?

1 Answer
Jul 3, 2017

There's probably an error in this question... I get #-"0.5331 kJ/mol"#, which is about 1000 times smaller than it should be.


Since we are in a bomb calorimeter, a constant-volume calorimeter, we have from the first law of thermodynamics:

#q_V = DeltaU + cancel(PDeltaV)^(0)#,

where:

  • #q_V# is the heat flow at constant volume.
  • #DeltaU# is the change in internal energy.
  • The work done, #-PDeltaV#, is defined here as the system doing work with respect to the surroundings.

Since we know the change in temperature and the combined heat capacity (presumably in #"kJ/"^@ "C"#...), we can find the heat involved in the reaction:

#|q_V| = C_VDeltaT#

#= "10.4 kJ/"^@ "C" xx 40^@ "C"#

#=# #"416 kJ"#

I interpret the molar heat of combustion to be #DeltabarU_C#, rather than the change in enthalpy (since the process is at constant volume, and we aren't given the volume of the calorimeter or water).

#DeltabarU_C = q_V/n_(LR)#,

where #n_(LR)# is the mols of the limiting reactant, i.e. the methanol in the presence of oxygen.

Assuming complete combustion:

#2"CH"_3"OH"(g) + 3"O"_2(g) -> 4"H"_2"O"(g) + 2"CO"_2(g)#

The molar internal energy of combustion is then the heat released out from the reaction towards the surrounding water:

#color(blue)(DeltabarU_C) = -("416 kJ")/(25000 cancel("g CH"_3"OH")) xx (32.04 cancel("g CH"_3"OH"))/"1 mol"#

#=# #color(blue)(-"0.5331 kJ/mol")#

This is an absurdly low number, so there's something off with the data... Are you sure it isn't #"25.000 g"#? The heat of combustion is off by over 1000 times.