Question

1. Given the chemical reaction:

3C_(graphite) + 4H_2(g) = C₃H_8(g)

Calculate the ΔH_rxn using Hess’s Law and the following information:

C₃H_8(g) + 5O_2(g) = 3CO_2(g) + 4H₂O_(l) ;    ΔH = -2220 kJ/mol

C_(graphite) + O_2(g) = CO_2(g) ;   ΔH = -393.5 kJ/mol

H_2(g) + ½ O_2(g)   = H₂O_(l) ;   ΔH = -285.8 kJ/mol

Answer 1

We have to calculate the delta H of the following reaction

3C (graphite) + 4 H2 (g) --- > C3H8 (g)

We use given reactions and their delta H to get delta H for this reaction

C₃H_8(g) + 5O_2(g) = 3CO_2(g) + 4H₂O_(l) ;    ΔH = -2220 kJ/mol       …1

C_(graphite) + O_2(g) = CO_2(g) ;   ΔH = -393.5 kJ/mol       …2

H_2(g) + ½ O_2(g)   = H₂O_(l) ;   ΔH = -285.8 kJ/mol      … 3

We multiply reaction second by 3 , reverse reaction first and multiply the reaction third by 4 and add resulting reaction and their added delta H will be the final delta of the above reaction in the problem.

3CO_2(g) + 4H₂O_(l) =   C₃H_8(g) + 5O_2(g)     ΔH = +2220 kJ/mol      

4H_2(g) + 2O_2(g)   = 4 H₂O_(l) ;   ΔH = (4* -285.8 kJ/mol)

3 C_(graphite) + 3O_2(g) = 3 CO_2(g) ;   ΔH = (3*-393.5 kJ/mol)      

------------------------------------------------    

3C(graphite) + 4 H2 (g) --- > C3H8 (g)      Delta H = [2220 + ( 4*-285.5 ) + ( 3*-393.5) ]kJ/mol

                                                 Delta H = - 102.5 kJ/mol