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Chapter 12: Problem 91

For the reaction:$$3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)$$. \(K=1.8 \times 10^{-7}\) at a certain temperature. If at equilibrium \(\left[\mathrm{O}_{2}\right]=0.062 \mathrm{M},\) calculate the equilibrium \(\mathrm{O}_{3}\) concentration.

Short Answer

Expert verified
The equilibrium concentration of \(O_3\) is approximately \(3.26\times10^{-4}\mathrm{M}\).

Step by step solution

01

Write the expression for the equilibrium constant (K)

The expression for K can be obtained from the balanced equation. For a general equilibrium reaction, it can be written as:$$K=\frac{([C]^{c}[D]^{d})}{([A]^{a}[B]^{b})}$$In our case, the reaction is: $$3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{O}_{3}(g)$$. We can write the expression for K as:$$K=\frac{[\mathrm{O}_{3}]^{2}}{[\mathrm{O}_{2}]^{3}}$$
02

Substitute the given values

We are given that $$ K=1.8 \times 10^{-7}$$ and the equilibrium concentration of $$O_2$$ is $$0.062 \mathrm{M}$$. We can substitute these values into the expression for K we found in Step 1.$$1.8\times10^{-7}=\frac{[\mathrm{O}_{3}]^{2}}{(0.062)^{3}}$$
03

Solve for the concentration of $$O_3$$

Now, we need to solve for the equilibrium concentration of $$O_3$$. We will first simplify the given equation and then find the value for $$[\mathrm{O}_{3}]$$. $$[\mathrm{O}_{3}]^{2}=(1.8\times10^{-7})(0.062^{3})$$Now, take the square root of both sides to solve for $$[\mathrm{O}_{3}]$$.$$[\mathrm{O}_{3}]=\sqrt{(1.8\times10^{-7})(0.062^{3})}$$
04

Calculate the equilibrium concentration of $$O_3$$

Now, use a calculator to find the equilibrium concentration of $$[\mathrm{O}_{3}]$$.$$[\mathrm{O}_{3}]=\sqrt{(1.8\times10^{-7})(0.062^{3})} \approx 3.26\times10^{-4} \mathrm{M}$$ So, the equilibrium concentration of $$O_3$$ is $$3.26\times10^{-4}\mathrm{M}$$.

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Most popular questions from this chapter

Nitrogen gas \(\left(\mathrm{N}_{2}\right)\) reacts with hydrogen gas \(\left(\mathrm{H}_{2}\right)\) to form ammonia \(\left(\mathrm{NH}_{3}\right) .\) At \(200^{\circ} \mathrm{C}\) in a closed container, 1.00 atm of nitrogen gas is mixed with 2.00 atm of hydrogen gas. At equilibrium,the total pressure is 2.00 atm. Calculate the partial pressure of hydrogen gas at equilibrium, and calculate the \(K_{\mathrm{p}}\) value for this reaction.

Hydrogen for use in ammonia production is produced by the reaction $$ \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \stackrel{\text { Ni catalyst }}{=\frac{750^{\circ} \mathrm{C}}{7}} \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) $$ What will happen to a reaction mixture at equilibrium if a. \(\mathrm{H}_{2} \mathrm{O}(g)\) is removed? b. the temperature is increased (the reaction is endothermic)? c. an inert gas is added to a rigid reaction container? d. \(\mathrm{CO}(g)\) is removed? e. the volume of the container is tripled?

Consider the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g) .\) 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 \(\mathrm{B} ?^{\prime \prime}\) What do you tell your friend?

Which of the following statements is(are) true? Correct the false statement(s). a. When a reactant is added to a system at equilibrium at a given temperature, the reaction will shift right to reestablish equilibrium. b. When a product is added to a system at equilibrium at a given temperature, the value of \(K\) for the reaction will increase when equilibrium is reestablished. c. When temperature is increased for a reaction at equilibrium, the value of \(K\) for the reaction will increase. d. When the volume of a reaction container is increased for a system at equilibrium at a given temperature, the reaction will shift left to reestablish equilibrium. e. Addition of a catalyst (a substance that increases the speed of the reaction) has no effect on the equilibrium position.

For the reaction $$2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$,it is determined that, at equilibrium at a particular temperature, the concentrations are as follows: \([\mathrm{NO}(g)]=8.1 \times 10^{-3} \mathrm{M}\) \(\left[\mathrm{H}_{2}(g)\right]=4.1 \times 10^{-5} \mathrm{M},\left[\mathrm{N}_{2}(g)\right]=5.3 \times 10^{-2} \mathrm{M},\) and \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=\) \(2.9 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.
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