Answer:
The bond formation of C-Cl bond by applying the given data in question 10 is 325.4 kJ
Explanation:
From the data in question 10:
The equation for the given reaction can be expressed as:
[tex]CH_{4(g)} + 2Cl_{(g)} \to CH_2Cl_2_{(g)}+H_2_{(g)}[/tex]
The enthalpy of formation of the compounds are as follows:
Substance [tex]\Delta \ H^0_f \ (kJ/mol)[/tex]
[tex]CH_{4(g)}[/tex] - 74.8
[tex]Cl_{(g)}[/tex] 120.9
[tex]CH_2Cl_{2(g)}[/tex] -95.8
From the data obtained above, the enthalpy of the reaction [tex]\Delta \ H ^0_{rxn}[/tex] can be computed as follows:
[tex]\Delta H^0_{rxn } = \sum ( \Delta H^0_f )_{products} - \sum ( \Delta H^0_f )_{reactants}[/tex]
[tex]\Delta H^0_{rxn } = [ 1 \ mol \times ( \Delta H^0_{CH_2Cl_{2(g)}}) + 1 \ mol \times ( \Delta H^0_{H_{2(g)}})] - [ 1 mol \times (\Delta H^0_{CH_4}_{(g)} +2 \ mol \times (\Delta \ H^0 _{Cl(g)})]}[/tex]
[tex]\Delta^0_{rxn } = [ 1 \ mol \times ( -95.8 \ kJ/mol) + 1 \ mol \times ( 0 \ kJ/mol] - [ 1 mol \times (-74.8 \ kJ/mol +2 \ mol \times (120.9 \ kJ/mol)]}[/tex]
[tex]\Delta H^0_{rxn } =-262.8 \ kJ[/tex]
However; since we knew [tex]\Delta H^0_{rxn }[/tex] to be -262.8 kJ
At standard conditions
Bond Bond Energy (kJ/mol)
C-H 410
H-H 432
The bond energy can be calculated by using the expression:
[tex]\Delta H^0_{rxn } = \sum \Delta H _{bond \ broken }- \sum \Delta H _{bond \ formed}[/tex]
[tex]-262. 8 KJ =[4 \times ( \Delta H_{CH})] - 1 [ (2 \times \Delta H_{CH} ) + ( 2 \times \Delta H_{C-Cl}) + (1 + \Delta H _{H-H})][/tex]
[tex]-262. 8 KJ =[4 \times ( 410 \ kJ/mol) ] - 1 [ (2 \times 410 \ kJ/mol ) + ( 2 \times \Delta H_{C-Cl}) + (1 +432 \ kJ/mol][/tex]
[tex]-262. 8 KJ = 388 \ kJ - ( 2 \times \Delta H_{C-Cl})[/tex]
[tex]( 2 \times \Delta H_{C-Cl}) = 388 \ kJ +262. 8 KJ[/tex]
[tex]( 2 \times \Delta H_{C-Cl}) = 650.8 \ KJ[/tex]
[tex]\Delta H_{C-Cl} = \dfrac{650.8 \ KJ}{2}[/tex]
[tex]\Delta H_{C-Cl} = 325.4 \ kJ[/tex]
Thus; the bond formation of C-Cl bond by applying the given data in question 10 is 325.4 kJ
