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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 for the given reaction are: \(\mathrm{SO}_{2}\): \(0.545M\), \(\mathrm{NO}_{2}\): \(0.545M\), \(\mathrm{SO}_{3}\): \(1.055M\), and \(\mathrm{NO}\): \(1.055M\).

Step by step solution

01

Write the equilibrium expression

To start, we need to write an equilibrium expression for the given chemical reaction. The equilibrium expression is the ratio of the product concentrations to the reactant concentrations. Each concentration is raised to the power of its stoichiometric coefficient. In this case, the equilibrium expression is: $$ K = \frac{[\mathrm{SO}_{3}][\mathrm{NO}]}{[\mathrm{SO}_{2}][\mathrm{NO}_{2}]} $$
02

Set up the ICE table

An ICE table (Initial, Change, Equilibrium) helps to organize the information in a systematic way. For this problem, the ICE table would look like: | | Initial | Change | Equilibrium | |:-:|---------|--------|-------------| | SO2 | 0.800 | -x | 0.800-x | | NO2 | 0.800 | -x | 0.800-x | | SO3 | 0.800 | +x | 0.800+x | | NO | 0.800 | +x | 0.800+x | The change in the concentrations is represented by "x". The reactants' concentrations decrease by x, while the products' concentrations increase by x.
03

Substitute equilibrium concentrations into the equilibrium expression

We now substitute the equilibrium concentrations into the equilibrium expression we derived in Step 1: $$ 3.75 = \frac{(0.800+x)(0.800+x)}{(0.800-x)(0.800-x)} $$
04

Solve for x

Next, we solve for x using quadratic equation or other algebraic methods. For the given equilibrium constant, the equation simplifies as: $$ 3.75 = \frac{(0.800+x)^{2}}{(0.800-x)^{2}} $$ Now, take the square root of both sides: $$ \sqrt{3.75} = \frac{0.800+x}{0.800-x} $$ $$ 1.935 = \frac{0.800+x}{0.800-x} $$ We can now cross multiply and solve for x: $$ 1.935(0.800-x)=(0.800+x) $$ $$ 1.548 - 1.935x = 0.800+x $$ $$ 2.935x = 0.748 $$ $$ x = 0.255 $$
05

Calculate the equilibrium concentrations

Finally, we can plug in x to calculate the equilibrium concentrations: $$ [\mathrm{SO}_{2}] = 0.800 - 0.255 = 0.545 M $$ $$ [\mathrm{NO}_{2}] = 0.800 - 0.255 = 0.545 M $$ $$ [\mathrm{SO}_{3}] = 0.800 + 0.255 = 1.055 M $$ $$ [\mathrm{NO}] = 0.800 + 0.255 = 1.055 M $$ Now, we have the equilibrium concentrations for all gases: SO2: 0.545 M NO2: 0.545 M SO3: 1.055 M NO: 1.055 M

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Most popular questions from this chapter

The hydrocarbon naphthalene was frequently used in mothballs until recently, when it was discovered that human inhalation of naphthalene vapors can lead to hemolytic anemia. Naphthalene is \(93.71 \%\) carbon by mass, and a \(0.256\) -mol sample of naphthalene has a mass of \(32.8 \mathrm{~g}\). What is the molecular formula of naphthalene? This compound works as a pesticide in mothballs by sublimation of the solid so that it fumigates enclosed spaces with its vapors according to the equation Naphthalene \((s) \rightleftharpoons\) naphthalene \((g)\) \(K=4.29 \times 10^{-6}(\) at \(298 \mathrm{~K})\) If \(3.00 \mathrm{~g}\) solid naphthalene is placed into an enclosed space with a volume of \(5.00 \mathrm{~L}\) at \(25^{\circ} \mathrm{C}\), what percentage of the naphthalene will have sublimed once equilibrium has been established?

Consider the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g) . \mathrm{A}\) friend asks the following: "I know we have been told that if a mixture of \(\mathrm{A}, \mathrm{B}, \mathrm{C}\), and \(\mathrm{D}\) is at equilibrium and more of \(\mathrm{A}\) is added, more \(\mathrm{C}\) and \(\mathrm{D}\) will form. But how can more \(\mathrm{C}\) and \(\mathrm{D}\) form if we do not add more B?" What do you tell your friend?

Consider the following reactions. \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g)\) and \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(s) \longrightarrow 2 \mathrm{HI}(g)\) List two property differences between these two reactions that relate to equilibrium.

For a typical equilibrium problem, the value of \(K\) and the initial reaction conditions are given for a specific reaction, and you are asked to calculate the equilibrium concentrations. Many of these calculations involve solving a quadratic or cubic equation. What can you do to avoid solving a quadratic or cubic equation and still come up with reasonable equilibrium concentrations?

Suppose a reaction has the equilibrium constant \(K=1.7 \times 10^{-8}\) at a particular temperature. Will there be a large or small amount of unreacted starting material present when this reaction reaches equilibrium? Is this reaction likely to be a good source of products at this temperature?
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