Question

consider the combustion reaction for salicylic acid (C7H6O3). the standard heat of formation of salicylic acid is deltaHf= +589.5 kj/mol. caclulate the amount of heat generated or released for the complete combustion of 4.85 g of C7H6O3.

Answer 1

combustion reaction of salicylic acid is

C₇H₆O₃ + 7O₂ --------> 7CO₂ + 3H₂O

H for this reaction is given by = H_f (products) - H_f (reactants)

H_f of C₇H₆O₃ = 589.5 KJ/mol

H_f of CO₂ = -393.5 KJ/mol

H_f of H₂O = -285.8 KJ/mol

H_f of O₂ = 0

So H_combustion = [ 7 * H_f (CO₂) + 3 * H_f (H₂O)] - [ H_f (C₇H₆O₃ ) + 7 * H_f (O₂) ]   

= [ 7 * (-393.5) + 3 * (-285.8) ] - [589.5 + 0 ]

= - 2754.5 - 857.4 - 589.5

  H_combustion    = -4201.4 KJ/mol

now weight of salicylic acid = 4.85 gm

mole of salicylic acid = 4.85 / 138 = 0.0351449 moles

so heat released by 0.0351449 moles = 4201.4 * 0.0351449

= 147.657 KJ

or we can write   H_combustion by 4.85 gm of salicylic acid is -147.657 KJ/mol (here - sign shows that heat is released.)