$$\ce{C(s) + 2 Cl2 (g) -> CCl4 (g)}$$ \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.

Using your Given data

$$2\times\Delta{}H^\circ_\mathrm{f} (\ce{Cl(g)}) +\Delta{}H^\circ_\mathrm{f} (\ce{C(g)}) -4\times\Delta H_{C-Cl}=\Delta{}H^\circ_\mathrm{f} (\ce{CCl4 (g)})$$

  • $\begingroup$ 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 $\Delta H_{C-Cl}$ $\endgroup$ – Baati Maliko Mar 20 '15 at 14:34
  • $\begingroup$ $\delta H$ 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. $\endgroup$ – Heisenberg Mar 20 '15 at 14:45
  • $\begingroup$ 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 . $\endgroup$ – Baati Maliko Mar 20 '15 at 15:03
  • $\begingroup$ You are taking account of energy here so $2$ is there because you have formed 2 moles of chlorine gas. $\endgroup$ – Heisenberg Mar 20 '15 at 15:06
  • $\begingroup$ but it says $2{} Cl_2$ That's Gonna Give Us $4 Cl$ if that's possible , because it says $\Delta{}H^\circ_\mathrm{f} (\ce{Cl(g)})$ $\endgroup$ – Baati Maliko Mar 20 '15 at 15:18

The bond dissociation enthalpy* $D$ is the enthalpy change for conversion of bonded into dissociated molecules following homolytic cleavage of the bond in question:

$$\ce{A-B (g)->A(g) + B(g)},\qquad D(\ce{A-B})=\Delta H^\circ_f (\ce{A(g)})+\Delta H^\circ_f (\ce{B(g)})-\Delta H^\circ_f (\ce{A-B(g)})$$

This is the property usually discussed as representing the "strength" of a bond, and is positive because the process of breaking a bond requires an input of heat (is endothermic). Breaking stronger bonds requires a greater positive enthalpy change.

The enthalpy change for formation of the bond by the dissociated molecular constituents is negative the dissociation enthalpy:

$$\Delta H^\circ (\ce{A-B})=-D (\ce{A-B})$$

and is a negative quantity because heat is released when bonds are formed (the process is exothermic).

The enthalpy of formation of a $\ce{Cl-Cl}$ bond is the enthalpy change for the following process:

$$\ce{2Cl(g)-> Cl2(g)}$$

and therefore can be expressed as

$$\Delta H^\circ (\ce{Cl-Cl})=\Delta H^\circ_f (\ce{Cl-Cl(g)})-2\Delta H^\circ_f (\ce{Cl(g)})$$

and since $\Delta H^\circ_f (\ce{Cl2(g)})=0$,

$$\Delta H^\circ (\ce{Cl-Cl})=-2\Delta H^\circ_f (\ce{Cl(g)})=-D(\ce{Cl-Cl})$$

In the process

$$\ce{C(g) + 4 Cl (g) -> CCl4 (g)}$$

four $\ce{C-Cl}$ bonds are formed, so the average heat of formation of one bond is

$$\Delta H^\circ(\ce{C-Cl}) = \frac{1}{4}\left(\Delta H^\circ_\mathrm{f}(\ce{CCl_4(g))}-\Delta H^\circ_\mathrm{f}(\ce{C(g)}) - 4 \Delta H^\circ_\mathrm{f}(\ce{Cl(g)}) \right) $$

which is a negative quantity (exothermic).

Note again that the corresponding bond dissociation enthalpy is then a positive quantity.

*Sources (from IUPAC Gold Book website):

  1. PAC, 1990, 62, 2167. (Glossary of atmospheric chemistry terms (Recommendations 1990)) on page 2176.
  2. PAC, 1994, 66, 1077. (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1089.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.