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Consider this reaction: $$\begin{array}{l}2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{CO}_{2}(g) \\ \Delta H=-1452.8 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ What is the value of \(\Delta H\) if (a) the equation is multiplied throughout by \(2 ;(b)\) the direction of the reaction is reversed so that the products become the reactants, and vice versa; (c) water vapor instead of liquid water is formed as the product?

Short Answer

Expert verified
(a) -2905.6 kJ/mol, (b) +1452.8 kJ/mol, (c) -1276.8 kJ/mol.

Step by step solution

01

Calculate ΔH when Equation is Multiplied by 2

When the chemical equation is multiplied by a factor of 2, the enthalpy change, \(\Delta H\), also gets multiplied by the same factor. The original \(\Delta H = -1452.8\, \text{kJ/mol}\). Multiply this by 2: \[ΔH = 2 \times (-1452.8\, \text{kJ/mol}) = -2905.6\, \text{kJ/mol}.\]
02

Calculate ΔH When Reaction Direction is Reversed

Reversing the direction of a chemical reaction changes the sign of \(\Delta H\) from negative to positive or vice versa. If the initial \(\Delta H\) is \(-1452.8\, \text{kJ/mol}\), the reversed reaction will have:\[ΔH = +1452.8\, \text{kJ/mol}.\]
03

Calculate ΔH for Formation of Water Vapor instead of Liquid Water

The standard enthalpy change of vaporization for water is \(\Delta H_v = 44.0 \text{kJ/mol}\). Since we have 4 moles of water being formed and turning into vapor instead of liquid, we need to adjust \(\Delta H\): \[ΔH = -1452.8\, \text{kJ/mol} + 4 \times 44.0\, \text{kJ/mol} = -1276.8\, \text{kJ/mol}.\]

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Key Concepts

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

Enthalpy Change
Enthalpy change, denoted by \( \Delta H \), represents the heat energy exchanged in a chemical reaction at constant pressure. It's a vital concept in thermochemistry, helping to determine whether a reaction releases or absorbs heat.
  • Exothermic reactions release heat, making \( \Delta H \) negative.
  • Endothermic reactions absorb heat, resulting in a positive \( \Delta H \).
Enthalpy change can be influenced by various factors, such as multiplying the reaction coefficients, reversing the reaction, or changing the physical state of products and reactants. Each of these scenarios alters \( \Delta H \) in a predictable way, showing how enthalpy change is essential in understanding energy transformations in reactions.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products and are accompanied by changes in energy. The given reaction involves two molecules of liquid methanol and three molecules of oxygen reacting to form water and carbon dioxide.
Governing these reactions are thermal dynamics principles like enthalpy change.
  • Reactions are influenced by temperature, pressure, and reaction conditions.
  • The stoichiometry, or the ratio of reactants to products, plays a crucial role in calculating enthalpy change.
Understanding these factors helps in predicting the behavior of chemical reactions, allowing for the manipulation of reaction conditions to achieve desired results. This fundamental knowledge is crucial for various applications, from industrial chemistry to studying natural processes.
Vaporization
Vaporization is the process where a liquid turns into vapor, which requires energy. This energy is known as the enthalpy of vaporization, \( \Delta H_v \), and it describes the amount of heat required to convert a substance from liquid to gas at constant pressure.
  • In the context of thermochemical problems, vaporization introduces an additional energy requirement.
  • For water, \( \Delta H_v = 44.0 \text{kJ/mol} \).
When a reaction outcome includes vapor instead of liquid water, the extra heat needed for vaporization must be added to the overall \( \Delta H \). This makes the reaction less exothermic because energy is consumed to vaporize the water, demonstrating the importance of state changes in measuring reaction energetics.
Reaction Direction Reversal
Reversing the direction of a chemical reaction is like pressing the rewind button. It flips the reactants to products and products to reactants. This reversal directly influences the enthalpy change.
  • When a reaction is reversed, the sign of the \( \Delta H \) changes. A negative \( \Delta H \) becomes positive, indicating the reaction absorbs energy.
  • Conversely, a positive \( \Delta H \) becomes negative, indicating a release of energy.
This principle helps chemists predict changes in energy requirements or generation when manipulating a reaction direction. It's a fundamental concept especially valuable in processes like chemical synthesis, where energy efficiency is crucial.

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Most popular questions from this chapter

What are the units for energy commonly employed in chemistry?

Consider two metals A and B, each having a mass of \(100 \mathrm{~g}\) and an initial temperature of \(20^{\circ} \mathrm{C}\). The specific heat of \(\mathrm{A}\) is larger than that of \(\mathrm{B}\). Under the same heating conditions, which metal would take longer to reach a temperature of \(21^{\circ} \mathrm{C} ?\)

Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and gasoline (assumed to be all octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\) ) are both used as automobile fuel. If gasoline is selling for \(\$ 2.20 / \mathrm{gal},\) what would the price of ethanol have to be in order to provide the same amount of heat per dollar? The density and \(\Delta H_{\mathrm{f}}^{\circ}\) of octane are \(0.7025 \mathrm{~g} / \mathrm{mL}\) and \(-249.9 \mathrm{~kJ} / \mathrm{mol}\), respectively, and of ethanol are \(0.7894 \mathrm{~g} / \mathrm{mL}\) and \(-277.0 \mathrm{~kJ} / \mathrm{mol}\) respectively \((1 \mathrm{gal}=3.785 \mathrm{~L})\).

Consider the reaction$$\begin{aligned}\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{HCl}(g) & \\\\\Delta H=-184.6 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$$If 3 moles of \(\mathrm{H}_{2}\) react with 3 moles of \(\mathrm{Cl}_{2}\) to form \(\mathrm{HCl}\) calculate the work done (in joules) against a pressure of \(1.0 \mathrm{~atm} .\) What is \(\Delta U\) for this reaction? Assume the reaction goes to completion and that \(\Delta V=0\). \((1 \mathrm{~L} \cdot \mathrm{atm}=101.3 \mathrm{~J})\)

Calculate \(q,\) and determine whether heat is absorbed or released when a system does work on the surroundings equal to \(64 \mathrm{~J}\) and \(\Delta U=213 \mathrm{~J}\).

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