given that:
delta H [Cl(g)]=121.3 kJ*mol^(-1)
delta H [C(g)]= 716.7 kJ*mol^(-1)
delta H [CCl(subscript 4)(g)]
=-95.7kJ*mol^(-1)
calculate the average molar bond enthalpy of the carbon-chlorine bond in a CCl(subscript 4) molecule.
The average molar bond enthalpy of the carbon-chlorine bond in a CCl₄ molecule can be calculated by using the given enthalpy values.
To find the average bond enthalpy, we need to consider the number of carbon-chlorine bonds in one CCl₄ molecule, which is 4.
The total change in enthalpy for breaking the 4 carbon-chlorine bonds in CCl₄ is:
ΔH(CCl₄) = 4 * ΔH(Cl(g)) - ΔH(C(g))
Substituting the given values:
ΔH(CCl₄) = 4 * 121.3 kJ*mol⁻¹ - 716.7 kJ*mol⁻¹
= 485.2 kJ*mol⁻¹ - 716.7 kJ*mol⁻¹
= -231.5 kJ*mol⁻¹
The negative sign indicates that energy is released when the bonds are broken.
To find the average molar bond enthalpy of carbon-chlorine bond, we divide the total change in enthalpy by the number of bonds:
Average bond enthalpy = ΔH(CCl₄) / Number of bonds
= -231.5 kJ*mol⁻¹ / 4
Therefore, the average molar bond enthalpy of the carbon-chlorine bond in a CCl₄ molecule is -57.9 kJ*mol⁻¹.
To calculate the average molar bond enthalpy of the carbon-chlorine bond in a CCl₄ molecule, we need to consider that CCl₄ contains four carbon-chlorine bonds.
The average molar bond enthalpy can be calculated by subtracting the sum of the bond enthalpies of the reactants from the sum of the bond enthalpies of the products:
Average molar bond enthalpy of C-Cl = [ΔH(CCl₄) - (ΔH(C) + 4ΔH(Cl))]/4
Now, let's plug in the values:
ΔH(CCl₄) = -95.7 kJ/mol
ΔH(C) = 716.7 kJ/mol
ΔH(Cl) = 121.3 kJ/mol
Average molar bond enthalpy of C-Cl = [-95.7 - (716.7 + 4 * 121.3)] / 4
First, calculate the value inside the parentheses:
= -95.7 - (716.7 + 485.2)
= -95.7 - 1201.9
= -1297.6 kJ/mol
Now, divide by 4 to get the average molar bond enthalpy:
Average molar bond enthalpy of C-Cl = -1297.6 / 4
= -324.4 kJ/mol
Therefore, the average molar bond enthalpy of the carbon-chlorine bond in a CCl₄ molecule is approximately -324.4 kJ/mol.