Question

A gaseous hydrocarbon reacts completely with oxygen gas to form carbon dioxide and water vapor. Given the following data, determine $\mathbf{∆}{\mathit{H}}_{\mathbf{f}}^{\mathbf{\circ }}$for the hydrocarbon:

$\begin{array}{rcl}\mathbf{∆}{\mathit{H}}_{\mathbf{r}\mathbf{x}\mathbf{n}}& \mathbf{=}& \mathbf{-}\mathbf{2044}\mathbf{.}\mathbf{5}\mathit{k}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\\ \mathbf{∆}{\mathit{H}}_{\mathbf{f}}^{\mathbf{\circ }}\mathbf{\left(}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{\right)}& \mathbf{=}& \mathbf{-}\mathbf{393}\mathbf{.}\mathbf{5}\mathit{k}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\\ \mathbf{∆}{\mathit{H}}_{\mathbf{f}}^{\mathbf{\circ }}\mathbf{\left(}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\right)}& \mathbf{=}& \mathbf{-}\mathbf{242}\mathit{k}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\end{array}$

Density of CO₂ and H₂O mixture at 1 atm, 200.^oC=0.751 g/L

The density of the hydrocarbon is less than the density of Kr at the same conditions.

Short answer

The $∆H_f^{\circ }$for the hydrocarbon is -490.75 kJmol^-1.

Step-by-step solution

Determination of molar mass of the hydrocarbon

The molar mass of hydrocarbon can be calculated by using the following formula,

PM = dRT

Where, P = pressure, d = density, R = gas constant, T = temperature

$$\begin{gathered} \mathit{M}\mathbf{=}\frac{\mathbf{d}\mathbf{R}\mathbf{T}}{\mathbf{P}} \\ \mathit{M}\mathbf{=}\frac{\left(0.751g/L\right)\left(0.0821Latm{K}^{-1}mo{l}^{-1}\right)\left(473K\right)}{\mathbf{1}\mathbf{a}\mathbf{t}\mathbf{m}} \\ \mathit{M}\mathbf{=}\mathbf{30}\mathit{G}\mathit{m}\mathit{o}{\mathit{l}}^{\mathbf{-}\mathbf{1}} \end{gathered}$$

Hence, the hydrocarbon can be C₂H₆.

Determination of ∆Hf∘ for hydrocarbon

The chemical reaction is,

$\begin{array}{rcl}\mathbf{2}{\mathit{C}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}\mathbf{+}\mathbf{7}{\mathit{O}}_{\mathbf{2}}& \mathbf{\to }& \mathbf{4}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{+}\mathbf{6}{\mathit{H}}_{\mathbf{2}}\mathit{O}\end{array}$

$\begin{array}{rcl}\mathbf{∆}{\mathit{H}}_{\mathbf{rxn}}^{\mathbf{\circ }}& \mathbf{=}& \mathbf{\sum }\mathbf{∆}{\mathit{H}}_{\mathbf{f}}^{\mathbf{\circ }}\mathbf{\left(}\mathit{p}\mathit{r}\mathit{o}\mathit{d}\mathit{u}\mathit{c}\mathit{t}\mathbf{\right)}\mathbf{-}\mathbf{\sum }\mathbf{∆}{\mathit{H}}_{\mathbf{f}}^{\mathbf{\circ }}\mathbf{\left(}\mathit{r}\mathit{e}\mathit{a}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\mathit{t}\mathbf{\right)}\\ \mathbf{-}\mathbf{2044}\mathbf{.}\mathbf{5}& \mathbf{=}& \mathbf{[}\left(4\times -393.5\right)\mathbf{+}\left(6\times -242\right)\mathbf{]}\mathbf{-}\mathbf{\left[}\left(2\times ∆H_f^{\circ }{C}_2{H}_6+0\right)\mathbf{\right]}\\ \mathbf{∆}{\mathit{H}}_{\mathbf{f}}^{\mathbf{\circ }}{\mathit{C}}_{\mathbf{2}}{\mathit{H}}_{\mathbf{6}}& \mathbf{=}& \mathbf{-}\mathbf{490}\mathbf{.}\mathbf{75}\mathit{k}\mathit{J}\mathit{m}\mathit{o}{\mathit{l}}^{\mathbf{-}\mathbf{1}}\end{array}$

Thus, the $\mathbf{∆}{\mathit{H}}_{\mathbf{f}}^{\mathbf{\circ }}$ for the hydrocarbon is -490.75 kJmol^-1.