A gaseous hydrocarbon reacts completely with oxygen gas to...

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

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

Key Concepts

Standard Enthalpy of Formation

This concept refers to the enthalpy change when one mole of a compound is formed from its elements in their standard states. It serves as a fundamental reference value in thermochemistry for calculating the heat changes of reactions, particularly when applying Hess’s Law.

Hess's Law

Hess's Law states that the overall enthalpy change for a chemical reaction is the same regardless of the pathway taken. This allows for the calculation of reaction enthalpies by summing the standard enthalpies of formation of products and subtracting those of the reactants.

Combustion Reaction Stoichiometry

Combustion reactions involve the complete reaction of a fuel, such as a hydrocarbon, with oxygen to form combustion products like carbon dioxide and water. Understanding the stoichiometry of these reactions is crucial for balancing chemical equations and relating the enthalpy changes to the molecular formulas of reactants and products.

Ideal Gas Law and Gas Density

The ideal gas law relates a gas's pressure, volume, temperature, and amount. It is a key tool for connecting measurements of gas density to the molar mass of the gas, thereby aiding in the determination of molecular formulas from physical properties.

Transcript

00:06 We're told that we have a gaseous hydrocarbon.

00:09 We don't know the composition of the hydrocarbon.

00:12 It reacts completely with oxygen gas to form carbon dioxide gas and water vapor.

00:26 We are to determine in this question here the enthalpy of formation of the hydrocarbon.

00:34 And we have some information given.

00:38 So let's use some variables here.

00:42 We use x to represent the moles of carbon and y to represent the moles of hydrogen.

00:50 And now let's balance this equation using these.

00:54 So i've already included x and y.

00:58 So let's modify our equation.

01:00 We have cx, h, y, gas, plus.

01:05 2x plus y over 2 over 2 oxygen plus x co2 and y h2 o't.

01:20 So with variables, there's our balanced equation.

01:25 Based on this, the enthalpy of the reaction is equal to the enthalpy of formation of the products minus enthalpy of formation of the reactants.

01:42 This would be equal to the infamy of formation.

01:48 This would be an x in here.

01:51 X, enthalpy of formation of co2, plus y.

02:01 Sorry, this should be y over 2.

02:04 Y over 2, enthalpy formation of h2, co2 gas, h2o gas minus the enthalpy formation cxhy plus the 2x plus a formation of o2 gas.

02:41 So let's plug in some of our numbers here.

02:44 We know that the enthalpy of the reaction is per mole of hydrocarbon carbon is negative 2 ,04 .5.

02:54 Will leave out the units kilojoules.

02:58 Enthopia formation of co2 gas is negative 393 .5 kilojoules plus y over two.

03:08 Enthopia formation of h2o gas is negative 242 kilojoules and this will be subtracted.

03:20 The entropy formation of cx, hy is unknown, plus the plus y over two over two of formation of o two gas is zero simplifying this we get negative 2 ,04 .5 is equal to negative 393 .5x minus 121y y minus delta h not formation cx, hy.

03:57 Now, let's use some of the information here...

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