Experimental data (simplified)

A bomb calorimeter had a heat capacity equivalent to 800.0 g of water.

After the complete combustion of 1.200 g of benzoic acid the temperature had risen by 9.500^oC.

This an adiabatic reaction, under these conditions because no heat is lost to the surroundings beyond the calorimeter and the volume remains constant, so it is ΔE (ΔU), the internal energy change, NOT ΔH.

 

Calculation

Heat energy given out to the water = mass of water x SHC_water x temperature change

q = m c ΔT (m in g, c = 44.18 Jg^-1K^-1, ΔT temperature rise in ^oC)

ΔE_calorimeter = 800.0 x 4.18 x 9.5 = 31768 J

M_r(C₆H₅COOH) = 122,  mol benzoic acid = 1.20 / 122 = 0.009836

Scaling up to 1 mole of benzoic acid: ΔE = 31768 x 1/0.009836 = 3229768 J

 

Correction to a standard enthalpy value

You now have to correct to non-adiabatic standard conditions using the formula

ΔH = ΔE + ΔnRT (Δn = the net change in moles of gas in the reaction)

The balanced equation under standard conditions is ...

C₆H₅COOH(s) +  7.5O₂(g)  ===>  7CO₂(g)  +  3H₂O(l)

ΔnRT = (7 – 7.5) x 8.314 x 298 = –1239 J

ΔH = ΔE + ΔnRT = 3229768 – 1239 = 3228529 J

ΔH^θ_comb.(C₆H₅COOH) = 3229 kJ mol^–1  (accurate to 4 sf)

 

Comments

This is a VERY accurate method and used to determine universal enthalpy values.

The accurate specific heat capacity of water (c, SHC) is actually 4.184 Jg^-1K^-1