Question

Use average bond energies to calculate ΔHrxn for the following hydrogenation reaction: H2C=CH2(g)+H2(g)→H3C−CH3(g)

Answer 1

Using average bond energies, the enthalpy of reaction for the hydrogenation of ethane is 2945.4 kJ/mol.

Let's consider the following hydrogenation reaction.

H₂C=CH₂(g) + H₂(g) → H₃C−CH₃(g)

Given the average bond energies (ΔH(bond)) we can calculate the enthalpy of the reaction (ΔHrxn) using the following equation.

$\Delta H_{rxn} = \Sigma \Delta H_{bonds\ broken} - \Sigma \Delta H_{bonds\ formed}$

The bonds broken, with their respective bond energies are:

H₂C=CH₂

• 1 C=C (615.0 kJ/mol)
• 4 C-H (416.2 kJ/mol)

H₂

• 1 H-H (435.1 kJ/mol)

The bonds formed, with their respective bond energies are:

H₃C−CH₃

• 1 C-C (347.3 kJ/mol)
• 6 C-H (416.2 kJ/mol)

The enthalpy of the reaction is:

$\Delta H _{rxn} = 6 \Delta H _{C=C} + 4 \Delta H _{C-H} + 1 \Delta H _{H-H} - 1 \Delta H _{C-C} - 6 \Delta H _{C-H}\\\\\Delta H _{rxn} = 6 (615.0 kJ/mol) + 1 (435.1 kJ/mol) - 1 (347.3kJ/mol) - 2 (416.2 kJ/mol) = 2945.4 kJ/mol$

Using average bond energies, the enthalpy of reaction for the hydrogenation of ethane is 2945.4 kJ/mol.

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Answer 2

The enthalpy of the hydrogenation reaction of ethane is 2945.4 kJ/mol.

The reaction is,

$\rm \bold{H_2C=CH_2(g)+H_2(g)\rightarrow H_3C-CH_3(g)}$

The bond energies of reactant and products are,

C=C (615.0 kJ/mol)

C-C (347.3 kJ/mol)

C-H (416.2 kJ/mol)

H-H (435.1 kJ/mol)

put this value in The Enthalpy formula,

$\rm \bold{ \Delta H _r_x_n = \sum \Delta H product - \Delta H reactant}\\\\\rm \bold{ \Delta H _r_x_n = (615.0 +4\times 416.2 + 435.1) -(347.3 + 6\times 416.2}\\\\\rm \bold{ \Delta H _r_x_n = 2945.4 kJ/mol. }$

So, we can conclude that the enthalpy of the given hydrogenation reaction of ethane is 2945.4 kJ/mol.

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