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Calculate the enthalpy change for the reaction $$ \mathrm{P}_{4} \mathrm{O}_{6}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) $$ given the following enthalpies of reaction: $$ \begin{array}{ll} \mathrm{P}_{4}(s)+3 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{6}(s) & \Delta H=-1640.1 \mathrm{~kJ} \\ \mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s) & \Delta H=-2940.1 \mathrm{~kJ} \end{array} $$

Short Answer

Expert verified
The enthalpy change for the given reaction, \(\mathrm{P}_{4}\mathrm{O}_{6}(s) + 2\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{10}(s)\), can be calculated by combining the given reactions and their enthalpy changes. The resulting enthalpy change for the target reaction is: $$ \Delta H_{\mathrm{overall}} = -1300\ \mathrm{kJ} $$

Step by step solution

01

Write down the given reactions and their enthalpies

We have the following reactions and their enthalpy changes: Reaction 1: $$ \mathrm{P}_{4}(s) + 3\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{6}(s) \\ \Delta H_1 = -1640.1\ \mathrm{kJ} $$ Reaction 2: $$ \mathrm{P}_{4}(s) + 5\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{10}(s) \\ \Delta H_2 = -2940.1\ \mathrm{kJ} $$ We want to calculate the enthalpy change for this reaction: $$ \mathrm{P}_{4}\mathrm{O}_{6}(s) + 2\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{10}(s) $$
02

Combine the reactions to form the target reaction

First, let's reverse Reaction 1 so that we have \(\mathrm{P}_{4}\mathrm{O}_{6}(s)\) on the reactant side: $$ \mathrm{P}_{4}\mathrm{O}_{6}(s) \longrightarrow \mathrm{P}_{4}(s) + 3\mathrm{O}_{2}(g) \\ \Delta H_1^\prime = 1640.1\ \mathrm{kJ} \qquad (\mathrm{since\ the\ reaction\ has\ been\ reversed}) $$ Now, let's add Reaction 1' and Reaction 2 together: Reaction 1': $$ \mathrm{P}_{4}\mathrm{O}_{6}(s) \longrightarrow \mathrm{P}_{4}(s) + 3\mathrm{O}_{2}(g) $$ Reaction 2: $$ \mathrm{P}_{4}(s) + 5\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{10}(s) $$ As we add the two reactions, \(\mathrm{P}_{4}(s)\) on the right of Reaction 1' cancels with \(\mathrm{P}_{4}(s)\) on the left of Reaction 2. Similarly, \(3\mathrm{O}_{2}(g)\) on the right of Reaction 1' combines with \(5\mathrm{O}_{2}(g)\) on the left of Reaction 2, resulting in \(2\mathrm{O}_{2}(g)\) on the left side of the overall reaction. Resulting overall reaction: $$ \mathrm{P}_{4}\mathrm{O}_{6}(s) + 2\mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4}\mathrm{O}_{10}(s) $$
03

Calculate the enthalpy change for the target reaction

As we add Reaction 1' and Reaction 2 to obtain the resulting overall reaction, we should also sum their enthalpy changes to obtain the enthalpy change for the target reaction. \(\Delta H_{\mathrm{overall}} = \Delta H_1^\prime + \Delta H_2 = 1640.1\ \mathrm{kJ} - 2940.1\ \mathrm{kJ} = -1300\ \mathrm{kJ}\) So, the enthalpy change for the target reaction is: $$ \Delta H_{\mathrm{overall}} = -1300\ \mathrm{kJ} $$

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Key Concepts

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

Hess's Law
Hess's Law is a fundamental concept in chemistry that provides a simple method for calculating reaction enthalpies. It states that if a reaction can be expressed as a series of steps, the total enthalpy change for the reaction equals the sum of the enthalpy changes for each step. This is possible because enthalpy is a state function, meaning it depends only on the initial and final states, not on the path taken.

In simpler terms, you can rearrange and add different chemical reactions to arrive at the reaction you're interested in, and then sum their enthalpy changes to find the overall enthalpy change.

The use of Hess's Law is particularly helpful when the direct enthalpy change of a reaction is difficult to measure. It allows chemists to use known enthalpy changes of related reactions to deduce the unknown enthalpy change, thus facilitating complex calculations.
  • Applicable to any process where the path does not matter.
  • Greatly simplifies the study of thermochemical processes.
  • Essential for understanding energy conservation in reactions.
In our exercise, Hess's Law was applied by reversing and combining the provided reactions, leading to the calculation of the enthalpy change for the desired reaction.
Thermochemical Equations
Thermochemical equations are chemical equations that include the enthalpy change associated with the chemical reaction. Apart from showing the reactants and products, these equations also indicate how much heat energy is either absorbed or released during the reaction.

The enthalpy change is denoted by \(\Delta H\) and can be positive or negative. A negative \(\Delta H\) indicates an exothermic reaction, where heat is released. Conversely, a positive \(\Delta H\) signifies an endothermic reaction, where heat is absorbed.

Writing thermochemical equations allows for a complete understanding of energy changes in reactions. This is crucial in predicting the feasibility of a reaction and its temperature conditions:
  • Shows the link between physical states of reactants and products.
  • Enthalpy change must be specified with balanced equations.
  • Includes phase states like (s), (l), (g), which are vital for accurate enthalpy calculation.
Thermochemical equations, like those in the exercise, help in laying out the required steps for using Hess's Law efficiently.
Reaction Enthalpy Calculations
Reaction enthalpy calculations focus on determining the heat change associated with chemical reactions. This involves several steps and considerations, mostly revolving around combining and manipulating thermochemical equations.

These calculations are crucial for understanding the energy dynamics of chemical processes. For calculating reaction enthalpies, the following key steps are usually involved:
  • Identify all given reactions and their enthalpy changes.
  • If needed, reverse or multiply reactions to match the desired target reaction.
  • Sum up the enthalpy changes of all the adjusted reactions.
  • Ensure that the final equation is balanced with respect to atoms and phases.
In the given exercise, reverse manipulation of one of the reactions and summing the enthalpy changes accurately yielded the desired target reaction enthalpy. This process showcases how reaction enthalpy calculations allow chemists to accurately estimate the heat changes for reactions that might not be directly measurable.

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Most popular questions from this chapter

(a) A serving of a particular ready-to-serve brown \& wild rice meal contains \(4.5 \mathrm{~g}\) fat, \(42 \mathrm{~g}\) carbohydrate, and \(4.0 \mathrm{~g}\) protein. Estimate the number of calories in a serving. (b) According to its nutrition label, the same meal also contains \(140 \mathrm{mg}\) of potassium ions. Do you think the potassium contributes to the caloric content of the food?

Without referring to tables, predict which of the following has the higher enthalpy in each case: (a) \(1 \mathrm{~mol} \mathrm{I}_{2}(s)\) or \(1 \mathrm{~mol} \mathrm{I}_{2}(g)\) at the same temperature, (b) \(2 \mathrm{~mol}\) of iodine atoms or \(1 \mathrm{~mol}\) of \(\mathrm{I}_{2},(\mathbf{c}) 1 \mathrm{~mol} \mathrm{I}_{2}(g)\) and \(1 \mathrm{~mol} \mathrm{H}_{2}(g)\) at \(25^{\circ} \mathrm{C}\) or \(2 \mathrm{~mol} \mathrm{HI}(g)\) at \(25^{\circ} \mathrm{C},(\mathbf{d}) 1 \mathrm{~mol} \mathrm{H}_{2}(g)\) at \(100^{\circ} \mathrm{C}\) or \(1 \mathrm{~mol} \mathrm{H}_{2}(g)\) at \(300^{\circ} \mathrm{C}\).

When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \operatorname{AgCl}(s) \quad \Delta H=-65.5 \mathrm{~kJ} $$ (a) Calculate \(\Delta H\) for the production of \(0.450 \mathrm{~mol}\) of \(\mathrm{AgCl}\) by this reaction. (b) Calculate \(\Delta H\) for the production of \(9.00 \mathrm{~g}\) of \(\mathrm{AgCl} . (\mathbf{c})\) Calculate \(\Delta H\) when \(9.25 \times 10^{-4} \mathrm{~mol}\) of \(\mathrm{AgCl}\) dissolves in water.

(a) What is meant by the term fuel value? (b) Which is a greater source of energy as food, \(5 \mathrm{~g}\) of fat or \(9 \mathrm{~g}\) of carbohydrate? (c) The metabolism of glucose produces \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). How does the human body expel these reaction products?

Under constant-volume conditions, the heat of combustion of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) is \(16.49 \mathrm{~kJ} / \mathrm{g}\). A \(3.00-\mathrm{g}\) sample of sucrose is burned in a bomb calorimeter. The temperature of the calorimeter increases from 21.94 to \(24.62^{\circ} \mathrm{C} .(\mathbf{a})\) What is the total heat capacity of the calorimeter? (b) If the size of the sucrose sample had been exactly twice as large, what would the temperature change of the calorimeter have been?

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