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Chapter 13: Problem 51

At a particular temperature, \(K=3.75\) for the reaction $$ \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g) $$ If all four gases had initial concentrations of \(0.800 M\), calculate the equilibrium concentrations of the gases.

Short Answer

Expert verified
The equilibrium concentrations of the gases are: \([\mathrm{SO}_2] = 0.593 \, M\), \([\mathrm{NO}_2] = 0.593 \, M\), \([\mathrm{SO}_3] = 1.007 \, M\), and \([\mathrm{NO}] = 1.007 \, M\).

Step by step solution

01

Interpret the reaction and identify initial concentrations

The chemical reaction given is: \[ \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g) \] The initial concentrations of each of the gases are 0.800 M, which can be represented as: - \([\mathrm{SO}_2]_0 = 0.800 \, M\) - \([\mathrm{NO}_2]_0 = 0.800 \, M\) - \([\mathrm{SO}_3]_0 = 0.800 \, M\) - \([\mathrm{NO}]_0 = 0.800 \, M\)
02

Define the changes in concentrations and equilibrium concentrations

Let's define the changes in concentrations at equilibrium as follows: - \([\mathrm{SO}_2] = 0.800 - x\, M\) - \([\mathrm{NO}_2] = 0.800 - x\, M\) - \([\mathrm{SO}_3] = 0.800 + x\, M\) - \([\mathrm{NO}] = 0.800 + x\, M\) The variables represent the change in concentrations during the reaction, where x is the amount of substance involved in the reaction. For example, if x = 0.200, that means 0.200 moles of the reactants have reacted to form 0.200 moles of the products.
03

Write the mass action expression for the reaction

Using the equilibrium constant K and the equilibrium concentrations, the mass action expression for this reaction is: \[ K = \frac{[\mathrm{SO}_{3}][\mathrm{NO}]}{[\mathrm{SO}_{2}][\mathrm{NO}_{2}]} \]
04

Substitute the known values and solve for the variable x

We're given K = 3.75 and the equilibrium concentrations, so substitute the values into the mass action expression: \[ 3.75 = \frac{(0.800 + x)(0.800 + x)}{(0.800 - x)(0.800 - x)} \] Now, solve the equation for x: \[ x^2 + 2(0.800)x - (0.800)^2 + \frac{(0.800)^2}{3.75} = 0 \] Solving this quadratic equation, we get: \[ x = 0.207\] Since we're dealing with concentrations and the quadratic formula may give us both positive and negative values, we only consider the positive value: x = 0.207.
05

Calculate the equilibrium concentrations of the gases

Use the value of x to calculate the equilibrium concentrations of the gases: - \([\mathrm{SO}_2] = 0.800 - x = 0.800 - 0.207 = 0.593 \, M\) - \([\mathrm{NO}_2] = 0.800 - x = 0.800 - 0.207 = 0.593 \, M\) - \([\mathrm{SO}_3] = 0.800 + x = 0.800 + 0.207 = 1.007 \, M\) - \([\mathrm{NO}] = 0.800 + x = 0.800 + 0.207 = 1.007 \, M\) Therefore, the equilibrium concentrations of the gases are: - \([\mathrm{SO}_2] = 0.593 \, M\) - \([\mathrm{NO}_2] = 0.593 \, M\) - \([\mathrm{SO}_3] = 1.007 \, M\) - \([\mathrm{NO}] = 1.007 \, M\)

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Most popular questions from this chapter

At high temperatures, elemental nitrogen and oxygen react with each other to form nitrogen monoxide: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ Suppose the system is analyzed at a particular temperature, and the equilibrium concentrations are found to be \(\left[\mathrm{N}_{2}\right]=\) \(0.041 \mathrm{M},\left[\mathrm{O}_{2}\right]=0.0078 M\), and \([\mathrm{NO}]=4.7 \times 10^{-4} M .\) Calcu- late the value of \(K\) for the reaction.

Consider the decomposition equilibrium for dinitrogen pentoxide: $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ At a certain temperature and a total pressure of \(1.00 \mathrm{~atm}\), the \(\mathrm{N}_{2} \mathrm{O}_{5}\) is \(0.50 \%\) decomposed (by moles) at equilibrium. a. If the volume is increased by a factor of \(10.0\), will the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposed at equilibrium be greater than, less than, or equal to \(0.50 \%\) ? Explain your answer. b. Calculate the mole percent of \(\mathrm{N}_{2} \mathrm{O}_{5}\) that will be decomposed at equilibrium if the volume is increased by a factor of \(10.0 .\)

Given the following equilibrium constants at \(427^{\circ} \mathrm{C}\), $$ \begin{array}{ll} \mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{1}=2 \times 10^{-25} \\ \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{2}=2 \times 10^{-5} \\ \mathrm{Na}_{2} \mathrm{O}_{2}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{3}=5 \times 10^{-29} \\ \mathrm{NaO}_{2}(s) \rightleftharpoons \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{4}=3 \times 10^{-14} \end{array} $$ determine the values for the equilibrium constants for the following reactions: a. \(\mathrm{Na}_{2} \mathrm{O}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) b. \(\mathrm{NaO}(\mathrm{g})+\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{Na}(l)\) c. \(2 \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) (Hint: When reaction equations are added, the equilibrium expressions are multiplied.)

A sample of gaseous nitrosyl bromide (NOBr) was placed in a container fitted with a frictionless, massless piston, where it decomposed at \(25^{\circ} \mathrm{C}\) according to the following equation: $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ The initial density of the system was recorded as \(4.495 \mathrm{~g} / \mathrm{L}\). After equilibrium was reached, the density was noted to be \(4.086 \mathrm{~g} / \mathrm{L}\) a. Determine the value of the equilibrium constant \(K\) for the reaction. b. If \(\operatorname{Ar}(g)\) is added to the system at equilibrium at constant temperature, what will happen to the equilibrium position? What happens to the value of \(K\) ? Explain each answer.

A gaseous material \(\mathrm{XY}(g)\) dissociates to some extent to produce \(\mathrm{X}(g)\) and \(\mathrm{Y}(g)\) : $$ \mathrm{XY}(g) \rightleftharpoons \mathrm{X}(g)+\mathrm{Y}(g) $$ A \(2.00-\mathrm{g}\) sample of \(\mathrm{XY}\) (molar mass \(=165 \mathrm{~g} / \mathrm{mol}\) ) is placed in a container with a movable piston at \(25^{\circ} \mathrm{C}\). The pressure is held constant at \(0.967\) atm. As \(X Y\) begins to dissociate, the piston moves until \(35.0\) mole percent of the original XY has dissociated and then remains at a constant position. Assuming ideal behavior, calculate the density of the gas in the container after the piston has stopped moving, and determine the value of \(K\) for this reaction of \(25^{\circ} \mathrm{C}\).
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