Molar Enthalpy Change

Question

A spirit burner was half filled with liquid heptane C7H16 weighed and put under a beaker containing 200cm3 of water in a tripod. When the temperature has risen exactly 20C the fire was extinguished and burner re-weighed. it had lost 0.5g in mass C7H16+11O2=7CO2=8H2O What is the molar enthalpy change ?

Answer 1

For this we have to assume that all of the energy from the combustion is transferred to the water. There are 200 cm3 of water, and the temperature goes up by 20° C. When heat is transferred we can calculate the amount of heat with the equation q = m•C•ΔT

m = mass of the sample in grams

C = specific heat capacity of the substance in Joules per gram-degree Celsius (J/g•°C)

ΔT = Temperature change (Tf - Ti usually but we are given that it went up by 20 degrees)

 

The beaker contains 200 cm³ of water, and the density of water is 1 g/cm³ so this is 200 g of water.

The specific heat capacity of water is 4.184 J/g•°C

 

So the heat absorbed by the water is

q = (200 g)(4.184 J/g•°C)(20° C) =  16736  J   (notice this is a + value, heat is absorbed by the water)

(also notice the mass units of grams and also the temperature units of degrees C cancel out here!) 

 

Not clear on how many sig figs you should have - technically only 1 given the data but I suspect that more are desired.  I'll go with 3 sig figs.

 

16700 J absorbed by the water, convert to kJ  (because most reaction enthalpy values are expressed in kJ/mol)

16700 J x 1 kJ/1000 J = 16.7 kJ

 

So we know that 16.7 kJ are absorbed by the water - where did it come from?  The burning of the heptane!  So the combustion reaction must have released 16.7 kJ, so the "q" of the reaction was -16.7 kJ (remember, exothermic processes have negative heat changes/enthalpies)

 

We can express the enthalpy change in terms of kJ per gram of heptane, but we are asked about the enthalpy of the reaction in kJ per mole of heptane.  All we have to do is convert grams to moles and get the ratio! We need the molecular weight of heptane, which is 100 g/mol

 

0.5 g of C7H16 x (1 mol/100 g)  = 0.005 mol

 

Enthalpy change of the reaction is therefore: -16.7 kJ/0.005 mol =  -3340 kJ/mol 

(don't forget that it is negative!) 

Usually we write it as "delta H"

ΔH = -3340 kJ/mol