Question

Calculate the enthalpy of formation of benzene given that enthalpy of combustion of benzene is - 3267.7

Answer 1

Given Thermochemical Equations:

(i) C6H6 + (15/2) O2 ----> 6CO2 + 3H2O ΔH = -3266.0 kJ

(ii) C + O2 ----> CO2 ΔH = -393.1 kJ

(iii) H2 + (1/2) O2 ---> H2O ΔH = -286.0 kJ


Now, we want enthalpy of formation of benzene, so we require C6H6 on the product side.

Reverse the equation (i); multiply the equation (ii) by 6 in order to get 6 CO2 ; multiply the equation (iii) by 3 to get 3 H2O. So, we will have:

(iv) 6CO2 + 3H2O ----> C6H6 + (15/2) O2 ΔH = + 3266.0 kJ

(v) 6C + 6O2 ----> 6CO2 ΔH = -2358.6 kJ

(vi) 3H2 + (3/2) O2 ---> 3H2O ΔH = -858.0 kJ

Notice that on reversing the equation (i) the sign of ΔH changes as well.

Now, observe that the 6CO​2 in reaction (iv) on the reactant side cancels out with 6CO​2 in reaction (v) in the product side.
Similarly, H2O and O2 also cancels out, as they are present in same molar quantities on both sides. So, we are left with:

6C + 3H2 ----> C6H6

and on adding their respective ΔH values we get the enthalpy of formation of benene = + 3266.0 kJ + (-2358.6 kJ) + (-858.0 kJ) = 49.4

Answer 2

Answer:

49.4

Explanation:

Given Thermochemical Equations:

(i) C6H6 + (15/2) O2 ----> 6CO2 + 3H2O ΔH = -3266.0 kJ

(ii) C + O2 ----> CO2 ΔH = -393.1 kJ

(iii) H2 + (1/2) O2 ---> H2O ΔH = -286.0 kJ

Now, we want enthalpy of formation of benzene, so we require C6H6 on the product side.

Reverse the equation (i);

multiply the equation (ii) by 6 in order to get 6 CO2 ;

multiply the equation (iii) by 3 to get 3 H2O.

So, we will have:

(iv) 6CO2 + 3H2O ----> C6H6 + (15/2) O2 ΔH = + 3266.0 kJ

(v) 6C + 6O2 ----> 6CO2 ΔH = -2358.6 kJ

(vi) 3H2 + (3/2) O2 ---> 3H2O ΔH = -858.0 kJ

Notice that on reversing the equation (i) the sign of ΔH changes as well.

Now, observe that the 6CO​2 in reaction (iv) on the reactant side cancels out with 6CO​2 in reaction (v) in the product side.

Similarly, H2O and O2 also cancels out, as they are present in same molar quantities on both sides. So, we are left with:

6C + 3H2 ----> C6H6

enthalpy of formation of benzene = + 3266.0 kJ + (-2358.6 kJ) + (-858.0 kJ)                                                                                                                                    =49.4