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Chapter 14: Problem 103

Phosphorus pentachloride gas decomposes to \(\mathrm{PCl}_{3}(\mathrm{~g})\) and \(\mathrm{Cl}_{2}(\mathrm{~g}) .\) At equilibrium, the concentrations of the decomposition products are \(5.50 \times 10^{-3} \mathrm{M}\) for \(\mathrm{PCl}_{3}(\mathrm{~g})\) and \(0.125 \mathrm{M}\) for \(\mathrm{Cl}_{2}(g)\). What is the equilibrium concentration of \(\mathrm{PCl}_{5}\) ? The equilibrium constant for this reaction is \(7.50 \times 10^{-2}\).

Short Answer

Expert verified
The equilibrium concentration of \(\mathrm{PCl}_{5}\) is approximately \(9.17 \times 10^{-3} \mathrm{M}\).

Step by step solution

01

Write the balanced chemical equation

The decomposition reaction of Phosphorus pentachloride (PCl5) is given by: \[\mathrm{PCl}_{5(g)} \rightleftharpoons \mathrm{PCl}_{3(g)} + \mathrm{Cl}_{2(g)}\]
02

Write the expression for the equilibrium constant

The equilibrium constant (K) for this reaction is related to the equilibrium concentrations of products and reactants by the following expression: \[K = \frac{[\mathrm{PCl}_3][\mathrm{Cl}_2]}{[\mathrm{PCl}_5]}\] Here, K, [\(\mathrm{PCl}_3\)], and [\(\mathrm{Cl}_2\)] are given, whereas [\(\mathrm{PCl}_5\)] is the unknown variable to solve.
03

Input the given equilibrium concentrations and equilibrium constant value

Now, we substitute the given values of K, [\(\mathrm{PCl}_3\)], and [\(\mathrm{Cl}_2\)] in the expression: \[7.50 \times 10^{-2} = \frac{(5.50 \times 10^{-3})(0.125)}{[\mathrm{PCl}_5]}\]
04

Solve for the unknown equilibrium concentration

We can rearrange the expression to find [\(\mathrm{PCl}_5\)]: \[[\mathrm{PCl}_5] = \frac{(5.50 \times 10^{-3})(0.125)}{7.50 \times 10^{-2}}\] After calculating, we get the equilibrium concentration of \(\mathrm{PCl}_{5(g)}\): \[[\mathrm{PCl}_5] \approx 9.17 \times 10^{-3} \mathrm{M}\] Therefore, the equilibrium concentration of \(\mathrm{PCl}_{5}\) is approximately \(9.17 \times 10^{-3} \mathrm{M}\).

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Most popular questions from this chapter

Sparingly soluble aluminum hydroxide dissolves in water to yield an equilibrium hydroxide ion concentration of \(8.58 \times 10^{-9} \mathrm{M}\). (a) Write the balanced equilibrium equation for aluminum hydroxide dissolving in water. (b) Write the \(K_{\mathrm{sp}}\) expression for aluminum hydroxide. (c) What is the equilibrium concentration of aluminum ion? (d) Calculate the value of \(K_{\mathrm{sp}}\) for aluminum hydroxide (show your calculation).

The saturation solubility of \(\mathrm{Ag}_{2} \mathrm{~S}\) at \(25^{\circ} \mathrm{C}\) is \(1.14 \times 10^{-17} \mathrm{M}\). What are the equilibrium concentrations of the cation and anion?

If \(k_{\mathrm{f}}>k_{\mathrm{r}}\), will \(K_{\mathrm{eq}}\) be less than 1 or greater than 1? Explain your answer.

Suppose the reaction \(\mathrm{A}_{2}+\mathrm{B}_{2} \rightleftarrows 2 \mathrm{AB}\) proceeds via a one-step mechanism involving a collision between one \(\mathrm{A}_{2}\) molecule and one \(\mathrm{B}_{2}\) molecule. Suppose also that this reaction is reversible, and that the forward reaction is inherently much faster than the reverse reaction. (a) Does the equilibrium lie to the left or to the right? Explain your choice in terms of the reactant and product concentrations necessary to establish equal forward and reverse rates. (b) Does an analysis in terms of the relationship \(K_{\mathrm{eq}}=k_{\mathrm{f}} / k_{\mathrm{r}}\) yield the same answer as in (a)? Explain.

At \(25^{\circ} \mathrm{C}\), the solubility of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in water is \(2.60 \times 10^{-6} \mathrm{M}\). What are the equilibrium concentrations of the cation and the anion in a saturated solution?
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