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Chapter 6: Problem 137

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:. $$ \begin{aligned} \Delta H_{\text {reaction }}^{\circ} &=-2044.5 \mathrm{~kJ} / \mathrm{mol} \text { hydrocarbon } \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{CO}_{2}\right) &=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right) &=-242 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ 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.

Short Answer

Expert verified
The enthalpy of formation for the gaseous hydrocarbon, \(\Delta H_f^\circ(C_xH_y)\), can be determined using Hess's Law, which results in the following equation: $$ \Delta H_f^\circ(C_xH_y) = \Delta H_f^\circ(CO_2) + \frac{y}{2x}\Delta H_f^\circ(H_2O) - \frac{\Delta H_\text{reaction}^\circ}{x} $$ With the given values, the equation becomes: $$ \Delta H_f^\circ(C_xH_y) = (-393.5\, kJ/mol) + \frac{y}{2x}(-242\, kJ/mol) - \frac{-2044.5\, kJ/mol}{x} $$ However, we cannot determine the exact enthalpy of formation without knowing the values of x and y.

Step by step solution

01

Write down the balanced chemical equation

Write the balanced chemical equation for the reaction of a gaseous hydrocarbon \((C_xH_y)\) with oxygen gas \((O_2)\) to form carbon dioxide \((CO_2)\) and water vapor \((H_2O)\): $$ xC_xH_y + yO_2 \rightarrow xCO_2 + \frac{y}{2}H_2O $$
02

Apply Hess's Law

Hess's Law states that the enthalpy change of the reaction is the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants. So we'll obtain the following definition: $$ \Delta H_\text{reaction}^\circ = \left[ x\Delta H_f^\circ(CO_2) + \frac{y}{2}\Delta H_f^\circ(H_2O) \right] - \left[ x\Delta H_f^\circ(C_xH_y) + 0 \right] $$ Note that the enthalpy of formation for elemental oxygen is zero because it is a pure element in its standard state.
03

Rearrange the equation to isolate ΔHf°(CxHy)

Rearrange the equation from Step 2 to isolate the enthalpy of formation of the gaseous hydrocarbon: $$ x\Delta H_f^\circ(C_xH_y) = x\Delta H_f^\circ(CO_2) + \frac{y}{2}\Delta H_f^\circ(H_2O) - \Delta H_\text{reaction}^\circ $$
04

Divide by x to obtain ΔHf°(CxHy)

Divide the equation from Step 3 by x to find the enthalpy of formation of the gaseous hydrocarbon per mole: $$ \Delta H_f^\circ(C_xH_y) = \Delta H_f^\circ(CO_2) + \frac{y}{2x}\Delta H_f^\circ(H_2O) - \frac{\Delta H_\text{reaction}^\circ}{x} $$
05

Plug in the given values and solve for ΔHf°(CxHy)

Now, plug in the given values for the different enthalpies of formation and the enthalpy of reaction to find the enthalpy of formation for the gaseous hydrocarbon: $$ \Delta H_f^\circ(C_xH_y) = (-393.5\, kJ/mol) + \frac{y}{2x}(-242\, kJ/mol) - \frac{-2044.5\, kJ/mol}{x} $$ Unfortunately, we don't have enough information to determine the exact enthalpy of formation for the hydrocarbon with this equation, since we don't know the values of x and y. However, this equation shows how to calculate the enthalpy of formation given the proper values for x and y.

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Most popular questions from this chapter

A coffee-cup calorimeter initially contains \(125 \mathrm{~g}\) water at \(24.2^{\circ} \mathrm{C}\). Potassium bromide \((10.5 \mathrm{~g})\), also at \(24.2^{\circ} \mathrm{C}\), is added to the water, and after the KBr dissolves, the final temperature is \(21.1^{\circ} \mathrm{C}\). Calculate the enthalpy change for dissolving the salt in \(\mathrm{J} / \mathrm{g}\) and \(\mathrm{kJ} / \mathrm{mol}\). Assume that the specific heat capacity of the solution is \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) and that no heat is transferred to the surroundings or to the calorimeter.

Given the following data $$ \begin{aligned} \mathrm{P}_{4}(s)+6 \mathrm{Cl}_{2}(g) & \longrightarrow 4 \mathrm{PCl}_{3}(g) & & \Delta H=-1225.6 \mathrm{~kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) & \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) & & \Delta H=-2967.3 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(g) & & \Delta H=-84.2 \mathrm{~kJ} \\ \mathrm{PCl}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Cl}_{3} \mathrm{PO}(g) & & \Delta H=-285.7 \mathrm{~kJ} \end{aligned} $$ calculate \(\Delta H\) for the reaction $$ \mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{PCl}_{5}(g) \longrightarrow 10 \mathrm{Cl}_{3} \mathrm{PO}(g) $$

Which of the following processes are exothermic? a. \(\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{~N}(g)\) b. \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(s)\) c. \(\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{Cl}(g)\) d. \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)\) e. \(\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{O}(g)\)

Consider \(2.00\) moles of an ideal gas that are taken from state \(A\) \(\left(P_{A}=2.00 \mathrm{~atm}, V_{A}=10.0 \mathrm{~L}\right)\) to state \(B\left(P_{B}=1.00 \mathrm{~atm}, V_{B}=\right.\) \(30.0 \mathrm{~L}\) ) by two different pathways: These pathways are summarized on the following graph of \(P\) versus \(V\) : Calculate the work (in units of J) associated with the two pathways. Is work a state function? Explain.

In a coffee-cup calorimeter, \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{AgNO}_{3}\) and \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) are mixed to yield the following reaction: $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \operatorname{AgCl}(s) $$ The two solutions were initially at \(22.60^{\circ} \mathrm{C}\), and the final temperature is \(23.40^{\circ} \mathrm{C}\). Calculate the heat that accompanies this reaction in \(\mathrm{kJ} / \mathrm{mol}\) of \(\mathrm{AgCl}\) formed. Assume that the combined solution has a mass of \(100.0 \mathrm{~g}\) and a specific heat capacity of \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g} .\)
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