Question

Use bond energies to determine δhrxn for the reaction between ethane and chlorine. ch3ch3(g)+cl2(g)→ch3ch2cl(g)+hcl(g)

Answer 1

The value of $\Delta {H_{{\text{reaction}}}}$ of the reaction ${\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_3}\left( g \right) + {\text{C}}{{\text{l}}_2}\left( g \right) \to {\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{Cl}}\left( g \right) + {\text{HCl}}\left( g \right)$ is $\boxed{ - 113\;{\text{kJ/mol}}}$.

Further explanation:

Heat of reaction:

The heat released or absorbed in a chemical reaction due to the difference in the bond energies (BE) of reactants and products in the reaction is known as the heat of reaction. It is represented by$\Delta {H_{{\text{reaction}}}}$.

The heat of reaction $\left( {\Delta {H_{{\text{reaction}}}}} \right)$can have two values:

Case I: If the reaction is endothermic, more energy needs to be supplied to the system than that released by it. So $\Delta {H_{{\text{reaction}}}}$ comes out to be positive.

Case II: If the reaction is exothermic, more energy is released by the system than that supplied to it. So $\Delta {H_{{\text{reaction}}}}$ comes out to be negative.

The formula to calculate the heat of reaction is,

$\boxed{\Delta {H_{{\text{reaction}}}} = \sum {\text{B}}{{\text{E}}_{{\text{reactant bond broken}}}} - \sum {\text{B}}{{\text{E}}_{{\text{product bond formed}}}}}$

Here,

$\Delta {H_{{\text{reaction}}}}$ is the heat of reaction.

${\text{B}}{{\text{E}}_{{\text{product bond formed}}}}$ is the bond energy of bond formation in products.

${\text{B}}{{\text{E}}_{{\text{reactant bond broken}}}}$ is the bond energy of bond breakage in reactants.

The given reaction occurs as follows:

${\text{C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_3}\left(g\right)+{\text{C}}{{\text{l}}_2}\left(g\right) \to {\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{Cl}}\left(g\right)+{\text{HCl}}\left(g\right)$

The number of broken bonds is one C-H bond and one Cl-Cl bond.

The number of bonds formed is one C-Cl bond and one H-Cl bond.

The formula to calculate the enthalpy of the given reaction is as follows:

$\Delta {H_{{\text{reaction}}}} = \left[ {\left( {1{\text{B}}{{\text{E}}_{{\text{C}} - {\text{H}}}} + 1{\text{B}}{{\text{E}}_{{\text{Cl}} - {\text{Cl}}}}} \right) - \left( {{\text{1B}}{{\text{E}}_{{\text{C}} - {\text{Cl}}}} + {\text{1B}}{{\text{E}}_{{\text{H}} - {\text{Cl}}}}} \right)} \right]$                …… (1)

The bond energy of C-H bond is 414 kJ/mol.

The bond energy of Cl-Cl bond is 243 kJ/mol.

The bond energy of C-Cl bond is 339 kJ/mol.

The bond energy of H-Cl bond is 431 kJ/mol.

Substitute these values in equation (1).

$\begin{aligned}\Delta {H_{{\text{reaction}}}}&=\left[ {\left( {41{\text{4 kJ/mol}}+243{\text{ kJ/mol}}} \right) - \left( {339{\text{ kJ/mol}}+431{\text{ kJ/mol}}}\right)}\right]\\&=\left[ {65{\text{7 kJ/mol}} - 77{\text{0 kJ/mol}}}\right]\\&=-11{\text{3 kJ/mol}}\\\end{aligned}$

Learn more:

1. Calculate the enthalpy change using Hess’s Law: brainly.com/question/11293201

2. Find the enthalpy of decomposition of 1 mole of MgO: brainly.com/question/2416245

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Thermodynamics

Keywords: C-Cl, C-H, H-Cl, Cl-Cl, -113 kJ/mol, heat of reaction, released, absorbed, exothermic, endothermic, positive, negative.

Answer 2

For this reaction to proceed, the following bond breaking should occur: 

*one C-H bond
* one Cl-Cl bond

After, the following bond formations should occur:
*one C-Cl bond
*one H-Cl bond

Now, add the bond energies for the respective bond energies which can be found in the attached picture. For bond formations, energy is negative. For bond breaking, energy is positive.

ΔHrxn = (1)(413) + (1)(242) + 1(-328) + 1(-431) = -104 kJ