Given the following information:
C(s) à C(g), H = +715 kJmol-1
Cl2(g) à 2Cl(g), H = +242 kJmol-1
C(s) + 2Cl2(g) à CCl4(g), H = -135.5 kJmol-1
Calculate the average bond dissociation enthalpy of a C-Cl bond.
To calculate the average bond dissociation enthalpy of a C-Cl bond, we need to use the information provided about the reactions involved.
The given reactions are:
1) C(s) → C(g), ΔH = +715 kJmol-1
2) Cl2(g) → 2Cl(g), ΔH = +242 kJmol-1
3) C(s) + 2Cl2(g) → CCl4(g), ΔH = -135.5 kJmol-1
We can see that the third reaction involves the formation of one C-Cl bond since CCl4 has four C-Cl bonds.
Now, let's consider the enthalpy change for the formation of one C-Cl bond:
C(s) + Cl2(g) → CCl(g) + Cl(g)
To calculate this enthalpy change, we can use the given reactions. We need to reverse and scale the second reaction, and then add it to the third reaction to cancel out the Cl(g) on the reactant side:
-1/2(Cl2(g) → 2Cl(g)) = -1/2 * (+242 kJmol-1) = -121 kJmol-1
Now, we add this result to the third reaction to obtain the enthalpy change for the formation of one C-Cl bond:
C(s) + Cl2(g) → CCl(g) + Cl(g) + 121 kJmol-1(Calculated by adding -121 kJmol-1)
Since we want to calculate the bond dissociation enthalpy of a C-Cl bond, we need to multiply the enthalpy change by -1 to reverse the reaction:
CCl(g) + Cl(g) → C(s) + Cl2(g) + 121 kJmol-1(Multiplied by -1)
Now, if we look at this equation, we can see that it represents the reverse of the first reaction. So, the enthalpy change for the formation of one C-Cl bond is:
ΔH = -715 kJmol-1
Therefore, the average bond dissociation enthalpy of a C-Cl bond is -715 kJmol-1.