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Chapter 19: Problem 64

Octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is a liquid hydrocarbon at room temperature that is the primary constituent of gasoline. (a) Write a balanced equation for the combustion of \(\mathrm{C}_{8} \mathrm{H}_{18}(l)\) to form \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) Without using thermochemical data, predict whether \(\Delta G^{\circ}\) for this reaction is more negative or less negative than \(\Delta H^{\circ}\).

Short Answer

Expert verified
The balanced chemical equation for the combustion of octane is: C8H18(l) + 12.5O2(g) -> 8CO2(g) + 9H2O(l) 鈭咷掳 for this reaction is more negative than 鈭咹掳, indicating the reaction is spontaneous at room temperature.

Step by step solution

01

Write the balanced chemical equation for the combustion of octane

To balance the chemical equation, we need to make sure that the number of carbon, hydrogen, and oxygen atoms in the reactants equals the number of those same atoms in the products. Reactants: C8H18(l) + O2(g) Products: CO2(g) + H2O(l) Start by balancing the carbons: C8H18(l) + O2(g) -> 8CO2(g) + H2O(l) Next, balance the hydrogens: C8H18(l) + O2(g) -> 8CO2(g) + 9H2O(l) Lastly, balance the oxygens: C8H18(l) + 12.5O2(g) -> 8CO2(g) + 9H2O(l) The balanced chemical equation for the combustion of octane is: C8H18(l) + 12.5O2(g) -> 8CO2(g) + 9H2O(l)
02

Predict if 鈭咷掳 is more negative or less negative than 鈭咹掳

We know that the relationship between Gibbs free energy (鈭咷掳), enthalpy (鈭咹掳), and entropy (鈭哠掳) is given by the equation: 鈭咷掳 = 鈭咹掳 - T鈭哠掳 In this case, we are not given any specific values for 鈭咹掳, 鈭哠掳, or T, so we cannot calculate the exact value of 鈭咷掳. However, we can still make predictions based on general trends. During the combustion of octane, a large amount of heat is released, which means that the reaction is exothermic (鈭咹掳 < 0). Moreover, the reaction has more gas molecules as products than reactants (8 moles of CO2 on the product side versus 12.5 moles of O2 on the reactant side), which leads to an increase in entropy (鈭哠掳 > 0). Since both 鈭咹掳 and 鈭哠掳 have negative and positive values, respectively, their product (-T鈭哠掳) will be negative. Therefore, 鈭咷掳 will be more negative than 鈭咹掳 in this reaction, indicating the reaction is spontaneous at room temperature. In conclusion, the balanced chemical equation for the given reaction is: C8H18(l) + 12.5O2(g) -> 8CO2(g) + 9H2O(l) And 鈭咷掳 for this reaction is more negative than 鈭咹掳.

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Key Concepts

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

Balanced Chemical Equation
A balanced chemical equation ensures that the number of each type of atom on the reactant side equals the number on the product side. When burning octane (\(\mathrm{C}_{8}\mathrm{H}_{18}(l)\)), you need to produce carbon dioxide and water.
Here's a simple way to balance it:
  • Count the carbon atoms in octane. You have 8, so you need 8 CO2.
  • Next, count the hydrogen atoms. There are 18, so you need 9 H2O, each H2O having 2 hydrogen atoms.
  • Finally, balance the oxygens. You'll end up needing 25 oxygen atoms, which comes from 12.5 O2, because 12.5 x 2 = 25.
So, the balanced equation is:\[\mathrm{C}_{8}\mathrm{H}_{18}(l) + 12.5\ \mathrm{O}_2(g) \rightarrow 8\ \mathrm{CO}_2(g) + 9\ \mathrm{H}_2\mathrm{O}(l)\]Balancing equations ensures that mass and charge are conserved, making them fundamental to chemical reactions.
Gibbs Free Energy
Gibbs Free Energy (\(\Delta G^{\circ}\)) tells us about the spontaneity of a reaction. It combines enthalpy (\(\Delta H^{\circ}\)), temperature (T), and entropy (\(\Delta S^{\circ}\)) in one equation:\[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\]A negative \(\Delta G^{\circ}\) means that the reaction can occur spontaneously.
In combustion, octane releases a lot of energy as heat, making it an exothermic reaction (\(\Delta H^{\circ} < 0\)). There鈥檚 also an increase in entropy because the reaction produces more gas molecules. The outcome is a more negative \(\Delta G^{\circ}\) than \(\Delta H^{\circ}\), reinforcing that the reaction is spontaneous.
Entropy
Entropy (\(\Delta S\)) is a measure of disorder or randomness in a system. When octane burns, the number of gas molecules increases from 12.5 (from O2) to 17 (8 CO2 and 9 H2O).
This transition results in higher entropy, as gases tend to have more disorder than liquids or solids.
During combustion:
  • You start with fewer gas molecules as reactants and end up with more as products.
  • This increase in randomness contributes positively to \(\Delta S\).
The greater the increase in entropy, the more it can drive a reaction to be spontaneous. This aligns well with the increase in entropy seen during the combustion of octane.

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Most popular questions from this chapter

Consider a reaction $\mathrm{A}_{2}(g)+\mathrm{B}_{2}(g) \rightleftharpoons 2 \mathrm{AB}(g),$ with atoms of A shown in red in the diagram and atoms of \(\mathrm{B}\) shown in blue. (a) If \(K_{\mathrm{c}}=1,\) which box represents the system at equilibrium? (b) If \(K_{\mathrm{c}}=1,\) which box represents the system at \(Q < K_{\mathrm{c}} ?(\mathbf{c})\) Rank the boxes in order of increasing magnitude of \(\Delta G\) for the reaction. [ Sections 19.5 and 19.7\(]\)

When most elastomeric polymers (e.g., a rubber band) are stretched, the molecules become more ordered, as illustrated here: Suppose you stretch a rubber band. (a) Do you expect the entropy of the system to increase or decrease? (b) If the rubber band were stretched isothermally, would heat need to be absorbed or emitted to maintain constant temperature? (c) Try this experiment: Stretch a rubber band and wait a moment. Then place the stretched rubber band on your upper lip, and let it return suddenly to its unstretched state (remember to keep holding on). What do you observe? Are your observations consistent with your answer to part (b)?

For the majority of the compounds listed in Appendix \(\mathrm{C},\) the value of \(\Delta G_{f}^{\circ}\) is more positive (or less negative) than the value of \(\Delta H_{f}^{\circ} .\) (a) Explain this observation, using \(\mathrm{NH}_{3}(g), \mathrm{CCl}_{4}(l)\), and \(\mathrm{KNO}_{3}(s)\) as examples. (b) An exception to this observation is \(\mathrm{CO}(g)\). Explain the trend in the \(\Delta H_{f}^{\circ}\) and \(\Delta G_{f}^{\circ}\) values for this molecule.

(a) Give two examples of endothermic processes that are spontaneous. (b) Give an example of a process that is spontaneous at one temperature but nonspontaneous at a different temperature.

(a) Using data in Appendix \(C\), estimate the temperature at which the free- energy change for the transformation from \(\mathrm{I}_{2}(s)\) to \(\mathrm{I}_{2}(g)\) is zero. What assumptions must you make in arriving at this estimate? (b) Use a reference source, such as Web Elements (www.webelements.com), to find the experimental melting and boiling points of \(\mathrm{I}_{2} .\) (c) Which of the values in part (b) is closer to the value you obtained in part (a)? Can you explain why this is so?
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