If for the reaction CCl_{4(g)} rightarrow C_{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The reaction is $CCl_{4(g)} \rightarrow C_{(g)} + 4Cl_{(g)}$.First,we find the enthalpy of formation of $CCl_{4(g)}$ using the enthalpy of vaporiz

Vedclass · Accepted Answer

$ 326 $

Vedclass · Answer

(B) The reaction is $CCl_{4(g)} ightarrow C_{(g)} + 4Cl_{(g)}$.First,we find the enthalpy of formation of $CCl_{4(g)}$ using the enthalpy of vaporization:$\Delta_{f} H^{	heta} (CCl_{4(g)}) = \Delta_{f} H^{	heta} (CCl_{4(l)}) + \Delta_{vap} H^{	heta} (CCl_{4(l)}) = -136.0 + 30.0 = -106.0 \ kJ \ mol^{-1}$.The enthalpy change for the reaction is given by:$\Delta_{r} H^{	heta} = \Delta_{a} H^{	heta} (C) + 4 	imes \Delta_{a} H^{	heta} (Cl) - \Delta_{f} H^{	heta} (CCl_{4(g)})$.Note that $\Delta_{a} H^{	heta} (Cl) = \frac{1}{2} \Delta_{a} H^{	heta} (Cl_2) = \frac{242.0}{2} = 121.0 \ kJ \ mol^{-1}$.$\Delta_{r} H^{	heta} = 714.0 + 4(121.0) - (-106.0) = 714.0 + 484.0 + 106.0 = 1304.0 \ kJ \ mol^{-1}$.The mean bond enthalpy of $C-Cl$ is $\frac{\Delta_{r} H^{	heta}}{4} = \frac{1304.0}{4} = 326 \ kJ \ mol^{-1}$.

Similar Questions

The heat of neutralization of the acid-base reaction is $57.32 \, kJ$ for which of the following pairs?

Hess's law of constant heat summation includes:

If the heats of combustion of $C$,$S$,and $CS_2$ are $x$,$y$,and $z \, kJ \, mol^{-1}$ respectively,what will be the heat of formation of $CS_2$?

The enthalpy of combustion of propane,graphite and dihydrogen at $298 \ K$ are: $-2220.0 \ kJ \ mol^{-1}$,$-393.5 \ kJ \ mol^{-1}$ and $-285.8 \ kJ \ mol^{-1}$ respectively. The magnitude of the enthalpy of formation of propane $(C_{3}H_{8})$ is ......... $kJ \ mol^{-1}$. (Nearest integer)

Vedclass Test Series

Exam Paper Generator

Online Exam Module