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Chapter 5: Problem 49

Gasohol, a mixture of ethyl alcohol and gasoline, has been proposed as a fuel to help conserve our petroleum resources. It is available on a limited basis. The thermochemical equation for the burning of ethyl alcohol is $$ \begin{aligned} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\ell)+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\ell) \\\ \Delta H=-1366.8 \mathrm{~kJ} \end{aligned} $$ Calculate the enthalpy change observed when burning \(2.00 \mathrm{~g}\) ethyl alcohol.

Short Answer

Expert verified
The enthalpy change is approximately -59.3 kJ.

Step by step solution

01

Find Molar Mass of Ethyl Alcohol

First, we need to find the molecular weight of ethyl alcohol (\(\text{C}_2\text{H}_5\text{OH}\)). The atomic masses are: Carbon (C) = 12.01 g/mol, Hydrogen (H) = 1.01 g/mol, and Oxygen (O) = 16.00 g/mol. Thus, the molar mass is \(2(12.01) + 6(1.01) + 16.00 = 46.08\, \text{g/mol}.\)
02

Calculate Moles of Ethyl Alcohol

To find the moles of ethyl alcohol in 2.00 g, use the formula: \(\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\). Thus, the number of moles is \(\frac{2.00\, \text{g}}{46.08\, \text{g/mol}} \approx 0.0434\, \text{mol}.\)
03

Use Thermochemical Equation

According to the thermochemical equation given, burning one mole of ethyl alcohol releases \(-1366.8\, \text{kJ}\). Therefore, to find the enthalpy change for 0.0434 moles, multiply the enthalpy change per mole by the number of moles: \((0.0434\, \text{mol}) \times (-1366.8\, \text{kJ/mol}) \approx -59.3\, \text{kJ}.\)

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Key Concepts

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

Enthalpy Change
Enthalpy change is a crucial concept in thermochemistry because it represents the heat absorbed or released during a chemical reaction under constant pressure. In our scenario, we look at the combustion of ethyl alcohol (or ethanol).
This reaction is exothermic, which means it releases heat to the surroundings, indicated by the negative sign of the enthalpy change: \(-1366.8\, \text{kJ/mol}\).
This value tells us how much heat is transferred when one mole of ethyl alcohol burns completely. By understanding the enthalpy change, we can gauge the energy efficiency of fuels when used in real-world applications such as gasohol in vehicles.
Molar Mass Calculation
Molar mass is the weight of one mole of a substance. It's an essential step in converting between grams and moles, which is vital for quantitative chemical analysis.
To calculate the molar mass of ethyl alcohol \(\text{C}_2\text{H}_5\text{OH}\), we have to sum up the atomic masses of all the atoms in the molecular formula.
For ethyl alcohol:
  • Carbon (C): \(2 \times 12.01\, \text{g/mol}\)
  • Hydrogen (H): \(6 \times 1.01\, \text{g/mol}\)
  • Oxygen (O): \(1 \times 16.00\, \text{g/mol}\)
Adding these gives us a molar mass of \(46.08\, \text{g/mol}\). This value becomes instrumental when calculating how many moles make up a specific mass of ethyl alcohol, such as the \(2.00\, \text{g}\) in our exercise.
Stoichiometry
Stoichiometry involves using mathematical relationships in balanced chemical equations to calculate the quantities of reactants and products involved.
In our enthalpy change problem, once we find the moles of ethyl alcohol using its molar mass, stoichiometry helps to determine the energy change. The balanced chemical equation tells us that for every mole of ethyl alcohol burnt, \(-1366.8\, \text{kJ}\) is released.
Therefore, if we burn \(0.0434\, \text{mol}\) of ethyl alcohol, the enthalpy change is calculated as:
\[0.0434\, \text{mol} \times -1366.8\, \text{kJ/mol} = -59.3\, \text{kJ}\]
This final calculation provides a quantitative understanding of the energy changes involved, allowing us to predict and measure reactions efficiently.

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Most popular questions from this chapter

What mass of ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\), must be burned to produce \(3420 \mathrm{~kJ}\) of heat, given that its enthalpy of combustion is \(-1410.1 \mathrm{~kJ} / \mathrm{mol} ?\)

Explain why absolute enthalpies cannot be measured and only changes can be determined.

The enthalpy change for the following reaction is \(+131.3 \mathrm{~kJ} .\) $$ \mathrm{C}(\mathrm{s} \text { , graphite })+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g}) $$ (a) Is energy released from or absorbed by the system in this reaction? (b) What quantities of reactants and products are $$ \text { assumed if } \Delta H=+131.3 \mathrm{~kJ} ? $$ (c) What is the enthalpy change when \(6.00 \mathrm{~g}\) carbon is reacted with excess \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) ?

Ammonia is produced commercially by the direct reaction of the elements. The formation of \(5.00 \mathrm{~g}\) gaseous \(\mathrm{NH}_{3}\) by this reaction releases \(13.56 \mathrm{~kJ}\) of heat. $$ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g}) $$ (a) What is the sign of the enthalpy change for this reaction? (b) Calculate \(\Delta H\) for the reaction, assuming molar amounts of reactants and products.

A \(50.0-\mathrm{g}\) sample of metal at \(100.00{ }^{\circ} \mathrm{C}\) is added to \(60.0 \mathrm{~g}\) water that is initially \(25.00{ }^{\circ} \mathrm{C}\). The final temperature of both the water and the metal is \(31.51^{\circ} \mathrm{C}\). (a) Use the specific heat of water to find the heat absorbed by the water. (b) How much heat did the metal sample lose? (c) Calculate the specific heat of the metal.
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