A gaseous hydrocarbon reacts completely with oxygen gas to form carbon dioxide and water vapor. Given the following data, determine \(\Delta H_{f}^{\circ}\) for the hydrocarbon: $$ \begin{aligned} \Delta H_{\mathrm{reacion}}^{\circ} &=-2044.5 \mathrm{kJ} / \mathrm{mol} \text { hydrocarbon } \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{CO}_{2}\right) &=-393.5 \mathrm{kJ} / \mathrm{mol} \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right) &=-242 \mathrm{kJ} / \mathrm{mol} \end{aligned} $$ Density of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) product mixture at 1.00 \(\mathrm{atm}\) , \(200 . \mathrm{C}=0.751 \mathrm{g} / \mathrm{L}\) . The density of the hydrocarbon is less than the density of Kr at the same conditions.

Step by step solution

01

Write the balanced chemical equation for the reaction.

Let the hydrocarbon be represented by \(C_xH_y\). The reaction is as follows: \[C_xH_y (g) + \left( x + \frac{y}{2} \right) O_2 (g) \rightarrow x CO_2 (g) + \frac{y}{2} H_2O(g)\]

02

Express the standard enthalpy change of the reaction in terms of the standard enthalpies of formation.

We know from Hess's law that: \[\Delta H_{reaction}^{\circ} = \sum n_i \Delta H_{f,i}^{\circ} (products) - \sum n_i \Delta H_{f,i}^{\circ} (reactants)\] From the balanced equation, substitute the values for the enthalpies of formation and the stoichiometric coefficients: \begin{align*} -2044.5\,\text{kJ/mol} &= x \left( -393.5 \,\text{kJ/mol} \right) + \frac{y}{2} \left( -242 \,\text{kJ/mol} \right) - \left( x + \frac{y}{2} \right) \Delta H_{f}^{\circ} (O_2) - \Delta H_{f}^{\circ} (C_xH_y) \\ -2044.5 &= -393.5x - 121y - \Delta H_{f}^{\circ} (C_xH_y) \end{align*} Since \(\Delta H_{f}^{\circ} (O_2) = 0\,\text{kJ/mol}\) for standard state oxygen gas.

03

Solve for the enthalpy of formation of the hydrocarbon.

We need to find the values of x and y for the hydrocarbon based on the given density information. However, without that information, we can still express the answer in terms of x and y as follows: \[\Delta H_{f}^{\circ} (C_xH_y) = -393.5x - 121y + 2044.5\] The standard enthalpy of formation for the gaseous hydrocarbon is expressed in terms of the number of carbon atoms (x) and hydrogen atoms (y). To find the precise value, further information about the hydrocarbon's composition is required.

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