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Hess's Law The enthalpy changes for the following reactions can be measured: \(\mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) $$ \Delta_{r} H^{\circ}=-802.4 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn} $$ \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})+3 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) \(\Delta_{1} H^{\circ}=-676 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn}\) (a) Use these values and Hess's law to determine the enthalpy change for the reaction \(\mathrm{CH}_{4}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) (b) Draw an energy level diagram that shows the relationship between the energy quantities involved in this problem.

Short Answer

Expert verified
The enthalpy change is \(-126.4 \text{ kJ/mol}\). The energy diagram shows a drop of 126.4 kJ from reactants to products.

Step by step solution

01

Understand the Known Reactions

We are given two known reactions with their enthalpy changes:1. \( \text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \) with \( \Delta H = -802.4 \text{ kJ/mol} \).2. \( \text{CH}_3\text{OH} + \frac{3}{2} \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \) with \( \Delta H = -676 \text{ kJ/mol} \).
02

Identify Target Reaction and Apply Hess's Law

The target is to find \( \Delta H \) for \( \text{CH}_4 + \frac{1}{2} \text{O}_2 \rightarrow \text{CH}_3\text{OH} \). According to Hess's Law, we can perform algebraic operations on the known reactions to achieve the target reaction.
03

Reverse the Second Reaction to Match Products with Target

Reverse the second reaction to form \( \text{CH}_3\text{OH} \) on the product side:1. \( \text{CO}_2 + 2 \text{H}_2\text{O} \rightarrow \text{CH}_3\text{OH} + \frac{3}{2} \text{O}_2 \) This reversal changes the sign of \( \Delta H \) to \(+676 \text{ kJ/mol} \).
04

Add Reactions to Obtain Target Reaction

Add the first reaction and the reversed second reaction:1. First: \( \text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2\text{O} \)2. Reversed second: \( \text{CO}_2 + 2 \text{H}_2\text{O} \rightarrow \text{CH}_3\text{OH} + \frac{3}{2} \text{O}_2 \)After canceling common terms, this results in: \( \text{CH}_4 + \frac{1}{2}\text{O}_2 \rightarrow \text{CH}_3\text{OH} \) with \( \Delta H = -802.4 + 676 = -126.4 \text{ kJ/mol} \).
05

Energy Level Diagram Explanation

Draw an energy level diagram with the energy levels of \( \text{CH}_4 + 2 \text{O}_2 \) and \( \text{CH}_3\text{OH} + \frac{3}{2} \text{O}_2 \) based on calculated and given \( \Delta H \) values. Indicate the transition from reactants to products showing the decrease in energy by 126.4 kJ when converting \( \text{CH}_4 + \frac{1}{2}\text{O}_2 \rightarrow \text{CH}_3\text{OH} \).

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Key Concepts

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

Enthalpy Change
The enthalpy change (\(\Delta H\)) of a reaction is the heat absorbed or released under constant pressure. It is a central concept in understanding energy transformations in chemical reactions.
Enthalpy describes the energy content of substances, and during a reaction, these content changes are observed as an enthalpy change. This change can be either:
  • Exothermic: where heat is released, and \(\Delta H\) is negative.
  • Endothermic: where heat is absorbed, and \(\Delta H\) is positive.
For instance, in the Hess's Law exercise provided, the enthalpy change for two reactions is given. The change must be calculated for a specific target reaction. Using Hess's Law, we can determine the overall enthalpy change by appropriately manipulating and combining known reactions.
To determine the enthalpy change for the reaction \(\text{CH}_4 + \frac{1}{2} \text{O}_2 \rightarrow \text{CH}_3\text{OH}\), we used the known reactions and applied algebraic manipulation processes to find that the enthalpy change is \(-126.4 \text{ kJ/mol}\). This indicates that when methane reacts with half a mole of oxygen to form methanol, \(126.4 \text{ kJ/mol}\) of energy are released.
Energy Level Diagram
An energy level diagram is a graphical representation of the energy changes during a chemical reaction.
It visually depicts the relative energy levels of reactants and products on a vertical scale, showing the energy pathway during the transformation.
This kind of diagram helps in understanding:
  • The energy level of reactants and products.
  • Whether a reaction is exothermic or endothermic.
  • The magnitude of energy change during the reaction, represented by the difference in height between reactants and products.
In the exercise, after calculating the enthalpy change, the energy level diagram was drawn to show this change. Reactants like \(\text{CH}_4 + 2 \text{O}_2\) start at a certain energy level, and the products \(\text{CH}_3\text{OH} + \frac{3}{2} \text{O}_2\) are at a lower level, representing the release of \(126.4 \text{ kJ/mol}\).
The diagram clearly illustrates the transition from high-energy reactants to lower-energy products, making it easier to comprehend the energy drop that occurs when methane and oxygen react to form methanol.
Chemical Reactions
In chemistry, a reaction occurs when substances (reactants) undergo changes to form new substances (products) with different properties.
Each reaction involves breaking bonds in reactants and forming bonds in products, accompanied by an energy change. This is why calculating enthalpy change and drawing energy level diagrams are important, they provide insight into the energy required or released during these transformations.
In the example provided, the target reaction involves methanol formation from methane and oxygen. This transformation is achieved by adjusting the known reactions using Hess's Law.
  • This reaction showcases how multiple steps in a chemical equation can be combined, highlighting Hess's Law's invaluable role in simplifying complex reactions to study net enthalpy changes.
  • It also illustrates how reversing reactions or changing coefficients can achieve a desired equation without experimental measurement, only theoretical calculation.
Understanding chemical reactions, energy changes, and how they interrelate allows us to predict behavior in extensive chemical processes, from industrial synthesis to environmental phenomena. This basis helps bridge theoretical chemistry with practical application.

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Most popular questions from this chapter

You want to heat the air in your house with natural gas \(\left(\mathrm{CH}_{4}\right) .\) Assume your house has \(275 \mathrm{m}^{2}\) (about \(2800 \mathrm{ft}^{2}\) ) of floor area and that the ceilings are \(2.50 \mathrm{m}\) from the floors. The air in the house has a molar heat capacity of \(29.1 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\). (The number of moles of air in the house can be found by assuming that the average molar mass of air is \(28.9 \mathrm{g} / \mathrm{mol}\) and that the density of air at these temperatures is \(1.22 \mathrm{g} / \mathrm{L} .\) ) What mass of methane do you have to burn to heat the air from \(15.0^{\circ} \mathrm{C}\) to \(22.0^{\circ} \mathrm{C} ?\)

A Three \(45-g\) ice cubes at \(0^{\circ} \mathrm{C}\) are dropped into \(5.00 \times 10^{2} \mathrm{mL}\) of tea to make iced tea. The tea was initially at \(20.0^{\circ} \mathrm{C} ;\) when thermal equilibrium was reached, the final temperature was \(0^{\circ} \mathrm{C}\) How much of the ice melted, and how much remained floating in the beverage? Assume the specific heat capacity of tea is the same as that of pure water.

Ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH},\) boils at \(78.29^{\circ} \mathrm{C} .\) How much energy, in joules, is required to raise the temperature of \(1.00 \mathrm{kg}\) of ethanol from \(20.0^{\circ} \mathrm{C}\) to the boiling point and then to change the liquid to vapor at that temperature? (The specific heat capacity of liquid ethanol is \(2.44 \mathrm{J} / \mathrm{g} \cdot \mathrm{K},\) and its enthalpy of vaporization is \(855 \mathrm{J} / \mathrm{g} .\) )

Suppose you burned 0.300 g of \(\mathrm{C}(\mathrm{s})\) in an excess of \(\mathbf{O}_{2}(g)\) in a constant-volume calorimeter to give \(\mathrm{CO}_{2}(\mathrm{g})\) $$ \mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}) $$ The temperature of the calorimeter, which contained 775 g of water, increased from \(25.00^{\circ} \mathrm{C}\) to \(27.38^{\circ} \mathrm{C} .\) The heat capacity of the bomb is \(893 \mathrm{J} / \mathrm{K} .\) Calculate \(\Delta U\) per mole of carbon.

A You are attending summer school and living in a very old dormitory. The day is oppressively hot, there is no air conditioner, and you can't open the windows of your room. There is a refrigerator in the room, however. In a stroke of genius, you open the door of the refrigerator, and cool air cascades out. The relief does not last long, though. Soon the refrigerator motor and condenser begin to run, and not long thereafter the room is hotter than it was before. Why did the room warm up?

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