# How to calculate the bond energy of C-Cl?

$$\ce{C(s) + 2 Cl2 (g) -> CCl4 (g)}$$ \begin{align} \Delta H^\circ_\mathrm{f} (\ce{C(g)}) &= 121~\mathrm{kJ/mol}\\ \Delta H^\circ_\mathrm{f} (\ce{Cl(g)}) &= 716~\mathrm{kJ/mol}\\ \Delta H^\circ_\mathrm{f} (\ce{CCl4(g)}) &= -106.48~\mathrm{kJ/mol}\\ \end{align}

Basically I already answered this, but I was a bit confused about the result, cause I don't know if there is a bond that is negative, cause am familiar with the energy given to "dissociation" of a bond or the "formation" of it , but here it's given the energy to the formation of chlorine and carbon, I don't know if that's the same as the formation of the bond or not!

\begin{align} \Delta H^\circ_\mathrm{f}(\ce{C-Cl}) &= \frac{\Delta H^\circ_\mathrm{f}(\ce{C(g)}) + 4 \Delta H^\circ_\mathrm{f}(\ce{Cl(g)}) - \Delta H^\circ_\mathrm{f}(\ce{CCl_4(g))}}{4} \\ \Delta H^\circ_\mathrm{f}(\ce{C-Cl}) &= \frac{121 + 4 \cdot 716 + 106.48}{4} \\ \Delta H^\circ_\mathrm{f}(\ce{C-Cl}) &= 772.87~\mathrm{kJ/mol} \end{align}

• Should I take the $\triangle H^\circ_\mathrm{f}(\ce{Cl(g)})$ same as $\triangle H^\circ_f(\ce{Cl-Cl})$? In that case I'm going to multiply it by $2$, not $4$.
• Should the $\Delta H^\circ_\mathrm{f}(\ce{C-Cl})$ be prefixed by a minus $-$ or a plus $+$, cause I don't know if it's an energy that is needed to formulate $\ce{CCl4}$? I'm not saying that the enthalpy of the bond should be negative, but the direction.

$$2\times\Delta{}H^\circ_\mathrm{f} (\ce{Cl(g)}) +\Delta{}H^\circ_\mathrm{f} (\ce{C(g)}) -4\times\Delta H_{C-Cl}=\Delta{}H^\circ_\mathrm{f} (\ce{CCl4 (g)})$$
• Hello Sir , thanks For Answering , Can you explain to me why You multiplied the $\triangle H^°_f(Cl(g))$ by $(2)$ , & Why You Put a $(-)$ Before The $\Delta H_{C-Cl}$ – Baati Maliko Mar 20 '15 at 14:34
• $\delta H$ depends on stoichiometry and I subtracted enthalpy of bond formation because that is the "lost" energy that is missing on RHS of equation .When bond is formed then energy is lost from LHS side into heat. Implying RHS has less energy than LHS so I just balanced out total energy. – Heisenberg Mar 20 '15 at 14:45
• that's seems Logical now , but still why $2\times\Delta{}H^\circ_\mathrm{f} (\ce{Cl(g)})$ . can we say That $2\times\Delta{}H^\circ_\mathrm{f} (\ce{Cl(g)})$ $= 1\triangle{}H^°_{d}{}_{(Cl-Cl)}$ , & thank you Sir for your time . – Baati Maliko Mar 20 '15 at 15:03
• You are taking account of energy here so $2$ is there because you have formed 2 moles of chlorine gas. – Heisenberg Mar 20 '15 at 15:06
• but it says $2{} Cl_2$ That's Gonna Give Us $4 Cl$ if that's possible , because it says $\Delta{}H^\circ_\mathrm{f} (\ce{Cl(g)})$ – Baati Maliko Mar 20 '15 at 15:18