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Chapter 13: Problem 95

Suppose \(K=4.5 \times 10^{-3}\) at a certain temperature for the reaction $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ If it is found that the concentration of \(\mathrm{PCl}_{5}\) is twice the concentration of \(\mathrm{PCl}_{3},\) what must be the concentration of \(\mathrm{Cl}_{2}\) under these conditions?

Short Answer

Expert verified
The concentration of $\mathrm{Cl}_{2}$ under these conditions is: $$[\mathrm{Cl}_{2}] = 9 \times 10^{-3} \mathrm{M}$$

Step by step solution

01

Write the equilibrium constant expression

For the given reaction: $$\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g) + \mathrm{Cl}_{2}(g)$$ We can write the equilibrium constant expression (K) as: $$K = \frac{[\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]}{[\mathrm{PCl}_{5}]}$$
02

Use the given information to create a relationship between the concentrations

We are informed that the concentration of PCl5 is twice the concentration of PCl3, which can be represented as: $$[\mathrm{PCl}_{5}] = 2\cdot [\mathrm{PCl}_{3}]$$
03

Replace the relationship in the equilibrium constant expression

Substitute the relationship between PCl5 and PCl3 concentrations in the equilibrium constant expression: $$K = \frac{[\mathrm{PCl}_{3}][\mathrm{Cl}_{2}]}{2\cdot[\mathrm{PCl}_{3}]}$$
04

Simplify and solve for the concentration of Cl2

Simplify the expression and solve for the concentration of Cl2: $$4.5 \times 10^{-3} = \frac{[\mathrm{Cl}_{2}]}{2}$$ Multiply both sides by 2 to isolate the concentration of Cl2: $$[\mathrm{Cl}_{2}] = 2 \times 4.5 \times 10^{-3}$$ Calculate the concentration of Cl2: $$[\mathrm{Cl}_{2}] = 9 \times 10^{-3} \mathrm{M}$$
05

Write the final answer

The concentration of Cl2 under these conditions is: $$[\mathrm{Cl}_{2}] = 9 \times 10^{-3} \mathrm{M}$$

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Most popular questions from this chapter

A sample of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is placed in an empty cylinder at \(25^{\circ} \mathrm{C}\) . After equilibrium is reached the total pressure is 1.5 atm and 16\(\%\) (by moles) of the original $\mathrm{N}_{2} \mathrm{O}_{4}(g)\( has dissociated to \)\mathrm{NO}_{2}(g) .$ a. Calculate the value of \(K_{\mathrm{p}}\) for this dissociation reaction at \(25^{\circ} \mathrm{C} .\) b. If the volume of the cylinder is increased until the total pressure is 1.0 atm (the temperature of the system remains constant), calculate the equilibrium pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(\mathrm{NO}_{2}(g) .\) c. What percentage (by moles) of the original $\mathrm{N}_{2} \mathrm{O}_{4}(g)$ is dissociated at the new equilibrium position (total pressure \(=1.00 \mathrm{atm} ) ?\)

Old-fashioned "smelling salts" consist of ammonium carbonate, (NH $_{4} )_{2} \mathrm{CO}_{3}$ . The reaction for the decomposition of ammonium carbonate $$\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ is endothermic. Would the smell of ammonia increase or decrease as the temperature is increased?

For the reaction $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2 \mathrm{CO}(g)$$ \(K_{\mathrm{p}}=2.00\) at some temperature. If this reaction at equilibrium has a total pressure of 6.00 \(\mathrm{atm}\) , determine the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the reaction container.

Explain why the development of a vapor pressure above a liquid in a closed container represents an equilibrium. What are the opposing processes? How do we recognize when the system has reached a state of equilibrium?

An 8.00 -g sample of \(\mathrm{SO}_{3}\) was placed in an evacuated container, where it decomposed at \(600^{\circ} \mathrm{C}\) according to the following reaction: $$\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)$$ At equilibrium the total pressure and the density of the gaseous mixture were 1.80 \(\mathrm{atm}\) and 1.60 \(\mathrm{g} / \mathrm{L}\) , respectively. Calculate \(K_{\mathrm{p}}\) for this reaction.
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