Question 9.4.1: Calculation of a Heat of Reaction from Heats of Combustion C......

Calculation of a Heat of Reaction from Heats of Combustion
Calculate the standard heat of reaction for the dehydrogenation of ethane:

C_{2}H_{6} → C_{2}H_{4} + H_{2}

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From Table B.1 or using APEx function DeltaHcg,

(\Delta \hat{H}^{°}_{c})_{ C_{2}H_{6}} = –  1559.9  kJ/mol\\ (\Delta \hat{H}^{°}_{c})_{ C_{2}H_{4}} = –  1411.0  kJ/mol\\ (\Delta \hat{H}^{°}_{c})_{ H_{2}} = –  285.84  kJ/mol

From Equation 9.4-1, therefore,

\begin{matrix}\boxed{ΔH^{°}_{r} = –  \sum\limits_{i}{\nu _{i}(\Delta \hat{H}^{°}_{c})_{i}} = \sum\limits_{reactants}{\left|\nu _{i}\right| (\Delta \hat{H}^{°}_{c})_{i}}- \sum\limits_{products}{\left|\nu _{i}\right| (\Delta \hat{H}^{°}_{c})_{i}}} & \pmb{(9.4-1)}\end{matrix}
\begin{matrix}ΔH^{°}_{r2} = 1  mol  C_{2}H_{6}(\Delta \hat{H}^{°}_{c})_{ C_{2}H_{6}}  –  1  mol  C_{2}H_{4}(\Delta \hat{H}^{°}_{c})_{ C_{2}H_{4}}  – 1  mol  H_{2} (\Delta \hat{H}^{°}_{c})_{ H_{2}} = \boxed{136.9  kJ}\end{matrix}

As an illustration, let us demonstrate the validity of this formula using Hess’s law. The combustion reactions are

1.  C_{2}H_{6} + \frac{7}{2}O_{2} → 2CO_{2} + 3H_{2}O
2.  C_{2}H_{4} + 3O_{2} → 2CO_{2} + 2H_{2}O
3.  H_{2} + \frac{1}{2} O_{2} → H_{2}O

It is easy to show that
4.  C_{2}H_{6} → C_{2}H_{4} + H_{2}

is obtained as (1) – (2) – (3). (Show it.) The desired result follows from Hess’s law.

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