Please send the answer of 8 th question.
Dear Student
This can be solved by Hess's law:
Combustion of benzene
C6H6 + 15/2O2 --------> 6CO2 + 3H2O ΔH = -3767.7 KJ/mol Eqn 1
Formation of CO2
C + O2 --------> CO2 ΔH = -393.5 KJ/mol Eqn 2
Formation of water
H2 + 1/2O2 --------> H2O ΔH = -286.6 KJ/mol Eqn 3
Now formation of benzene can be represented as:
6C + 3H2 -------> C6H6
On multiplying eqn 3 with 3 we get
3*(H2 + 1/2O2 --------> H2O ΔH = -286.6 KJ/mol )
3H2 + 3/2O2 --------> 3H2O ΔH = -859.8 KJ Eqn 4
Multiplying eqn 2 with 6, we get
6*( C + O2 --------> CO2 ΔH = -393.5 KJ/mol
6C + 6O2 --------> 6CO2 ΔH = -2361.0 KJ/mol Eqn 5
On adding eqn 4 and eqn 5, and then subtracting eqn 1 from the sum, we get
Subtracting Eqn 1 from eqn 5, we get equation for formation of methane,
6C + 3H2 -------> C6H6
ΔH =[-859.8 +(-2361)] - (-3767.7) = +546.2 KJ/mol
Regards
This can be solved by Hess's law:
Combustion of benzene
C6H6 + 15/2O2 --------> 6CO2 + 3H2O ΔH = -3767.7 KJ/mol Eqn 1
Formation of CO2
C + O2 --------> CO2 ΔH = -393.5 KJ/mol Eqn 2
Formation of water
H2 + 1/2O2 --------> H2O ΔH = -286.6 KJ/mol Eqn 3
Now formation of benzene can be represented as:
6C + 3H2 -------> C6H6
On multiplying eqn 3 with 3 we get
3*(H2 + 1/2O2 --------> H2O ΔH = -286.6 KJ/mol )
3H2 + 3/2O2 --------> 3H2O ΔH = -859.8 KJ Eqn 4
Multiplying eqn 2 with 6, we get
6*( C + O2 --------> CO2 ΔH = -393.5 KJ/mol
6C + 6O2 --------> 6CO2 ΔH = -2361.0 KJ/mol Eqn 5
On adding eqn 4 and eqn 5, and then subtracting eqn 1 from the sum, we get
Subtracting Eqn 1 from eqn 5, we get equation for formation of methane,
6C + 3H2 -------> C6H6
ΔH =[-859.8 +(-2361)] - (-3767.7) = +546.2 KJ/mol
Regards