/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 137 A gaseous hydrocarbon reacts com... [FREE SOLUTION] | 91Ó°ÊÓ

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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

enthalpy of formation
Enthalpy of formation, often represented as \( \Delta H_f^\circ \), is a fundamental aspect of thermochemistry that defines the heat change when one mole of a compound is formed from its elements in their standard states. Standard states refer to the most stable forms of each element at a pressure of 1 atmosphere and a specific temperature, usually 25 °C (298 K).
The enthalpy of formation is particularly useful because it allows for the determination of other enthalpy changes, such as the heat of reaction. By using known enthalpies of formation, you can apply Hess's Law to find the enthalpy change of any chemical reaction. This is achieved by comparing the enthalpy of formation of products and reactants, thereby predicting whether a reaction will release or absorb heat. It is crucial for understanding energy exchanges in various chemical processes.
Hess's Law
Hess's Law is a principle in thermochemistry that states the overall enthalpy change in a chemical reaction is the same, regardless of the number of steps taken to perform the reaction. This law relies on the concept that enthalpy, a state function, depends only on the initial and final states of the system, not the path taken.
To apply Hess's Law effectively, follow these steps:
  • Identify the target reaction and its enthalpy change that you want to calculate.
  • Use the enthalpy of formation values for each product and reactant from known data.
  • Structure the equation to express the total enthalpy change as the sum of the enthalpies of formation of products minus the reactants.
Following Hess's Law simplifies the estimation of reaction enthalpies, especially when direct measurement is challenging. It is a key tool in predicting energy requirements and expectations for chemical reactions in science and industry.
chemical equations
Chemical equations provide a concise way to represent a chemical reaction. They include reactants, products, and their respective quantities, often indicated in moles. Every chemical equation needs to be balanced to comply with the Law of Conservation of Mass, meaning the same number of each type of atom must be present on both sides of the equation.
In writing a balanced chemical equation for a hydrocarbon reacting with oxygen, for example, you'd ensure carbon atoms from the hydrocarbon appear as carbon dioxide, and hydrogen atoms as water. Oxygen from the reactants must also balance with the oxygen present in the products.
Balanced equations are not only integral for correct stoichiometric calculations but also to accurately determine the thermochemical quantities, like the enthalpy change in a reaction. Balancing equations is a critical skill for predicting product yields and understanding how much reactant is necessary for a reaction.
stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions. It is the mathematical relationship between the quantities used and formed, based on the balanced chemical equation.
Key components in stoichiometric calculations include:
  • Mole ratio: Derived from the coefficients of a balanced chemical equation.
  • Conversion factors: Used to convert between moles, mass, and volume of reactants/products.
  • Limiting reactant: The reactant that will run out first, determining the maximum amount of product formed.
  • Excess reactants: Reactants that remain after the reaction is complete.
Using stoichiometry, you can predict how much product will form from given reactants, or how much of each reactant is necessary to produce a desired amount of product. It's a powerful tool that connects chemical equation theory to real-world laboratory practice and industrial applications.

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Most popular questions from this chapter

The preparation of \(\mathrm{NO}_{2}(g)\) from \(\mathrm{N}_{2}(g)\) and \(\mathrm{O}_{2}(g)\) is an endothermic reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) \text { (unbalanced) } $$ The enthalpy change of reaction for the balanced equation (with lowest whole- number coefficients) is \(\Delta H=67.7 \mathrm{~kJ}\). If \(2.50 \times 10^{2} \mathrm{~mL} \mathrm{~N}_{2}(g)\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(3.50 \mathrm{~atm}\) and \(4.50 \times\) \(10^{2} \mathrm{~mL} \mathrm{O}_{2}(g)\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(3.50\) atm are mixed, what amount of heat is necessary to synthesize the maximum yield of \(\mathrm{NO}_{2}(g) ?\)

A swimming pool, \(10.0 \mathrm{~m}\) by \(4.0 \mathrm{~m}\), is filled with water to a depth of \(3.0 \mathrm{~m}\) at a temperature of \(20.2^{\circ} \mathrm{C}\). How much energy is required to raise the temperature of the water to \(24.6^{\circ} \mathrm{C} ?\)

In a coffee-cup calorimeter, \(150.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HCl}\) is added to \(50.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{NaOH}\) to make \(200.0 \mathrm{~g}\) solution at an initial temperature of \(48.2^{\circ} \mathrm{C}\). If the enthalpy of neutralization for the reaction between a strong acid and a strong base is \(-56 \mathrm{~kJ} / \mathrm{mol}\), calculate the final temperature of the calorimeter contents. Assume the specific heat capacity of the solution is \(4.184 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and assume no heat loss to the surroundings.

One way to lose weight is to exercise! Walking briskly at \(4.0\) miles per hour for an hour consumes about 400 kcal of energy. How many hours would you have to walk at \(4.0\) miles per hour to lose one pound of body fat? One gram of body fat is equivalent to \(7.7\) kcal of energy. There are \(454 \mathrm{~g}\) in \(1 \mathrm{lb}\).

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.

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