Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 \(9.17 \times 10^{-3} \mathrm{M}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
The equilibrium constant, often denoted as \( K \), is a mathematical expression that quantifies the ratio of the concentrations of the products to the reactants of a reaction at equilibrium. In this reaction, the equilibrium constant was given as \( 7.50 \times 10^{-2} \). It tells us about the position of equilibrium and whether the products or reactants are favored under a given set of conditions. If \( K > 1 \), the products are favored at equilibrium, meaning more products are present than reactants. If \( K < 1 \), the reactants are favored, indicating more reactants than products at equilibrium. In our example, with the decomposition of phosphorus pentachloride, the equilibrium constant is less than 1, suggesting that the formation of reactants is somewhat favored in this situation.
Concentration Calculations
Concentration calculations are essential for determining the amount of a substance present in a reaction at equilibrium. Here, we were given the equilibrium concentrations for the products: \[[\mathrm{PCl}_3] = 5.50 \times 10^{-3} \text{ M} \] and \[[\mathrm{Cl}_2] = 0.125 \text{ M} \].To find the concentration of \( \mathrm{PCl}_5 \), the missing piece, we rearranged the equilibrium expression to solve for it: \[[\mathrm{PCl}_5] = \frac{(5.50 \times 10^{-3})(0.125)}{7.50 \times 10^{-2}} \approx 9.17 \times 10^{-3} \text{ M} \]This calculation shows how products and reactants relate at equilibrium and how to manipulate equations to find unknown concentrations.
Balanced Chemical Equation
Creating a balanced chemical equation is the first critical step in any equilibrium calculation. This ensures that the law of conservation of mass is respected. It involves making sure that there are equal numbers of each type of atom on both sides of the equation.For phosphorus pentachloride (\( \mathrm{PCl}_5 \)) decomposition, the balanced reaction is:\[\mathrm{PCl}_{5(g)} \rightleftharpoons \mathrm{PCl}_{3(g)} + \mathrm{Cl}_{2(g)} \]This reaction shows that one molecule of \( \mathrm{PCl}_5 \) decomposes into one molecule each of \( \mathrm{PCl}_3 \) and \( \mathrm{Cl}_2 \). A balanced chemical equation is your starting point for setting up the equilibrium constant expression.
Phosphorus Pentachloride Decomposition
Phosphorus pentachloride (\( \mathrm{PCl}_5 \)) is known to decompose into phosphorus trichloride (\( \mathrm{PCl}_3 \)) and chlorine gas (\( \mathrm{Cl}_2 \)) under specific conditions. This reaction can be reversed, meaning that these products can also form \( \mathrm{PCl}_5 \) again, reaching an equilibrium state.The decomposition can be summarized by the equation:\[\mathrm{PCl}_{5(g)} \rightleftharpoons \mathrm{PCl}_{3(g)} + \mathrm{Cl}_{2(g)} \]This reversible reaction showcases the dynamic nature of chemical equilibria. Initially, only \( \mathrm{PCl}_5 \) might be present, but as it decomposes, \( \mathrm{PCl}_3 \) and \( \mathrm{Cl}_2 \) form until the rate of decomposition matches the rate of recombination. This results in a stable balance where the concentrations no longer change.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose a reaction is at equilibrium and you then disturb the equilibrium by adding reactants. What happens to the value of \(K_{\text {eq }}\) ? 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.

Sparingly soluble calcium phosphate dissolves in water to yield an equilibrium calcium ion concentration of \(7.8 \times 10^{-6} \mathrm{M}\). (a) Write the balanced equilibrium equation for calcium phosphate dissolving in water. (b) Write the \(K_{\text {sp }}\) expression for calcium phosphate. (c) What is the equilibrium concentration of phosphate ion? (d) Calculate the value of \(K_{s p}\) for calcium phosphate (show your calculation).

Suppose you have a reaction with many reactants. When you write the equilibrium expression for the reaction, do the reactant concentrations all go in the numerator or in the denominator? What mathematical operation(s) should be used for these concentrations?

The process of photosynthesis in plants converts carbon dioxide and water to glucose and oxygen: \(6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \quad \Delta E_{\mathrm{rxn}}=2801 \mathrm{~kJ}\) (a) Write the equilibrium constant expression for this conversion. (b) How would the equilibrium be affected if \(\mathrm{CO}_{2}(g)\) were added? (c) How would the equilibrium be affected if \(\mathrm{H}_{2} \mathrm{O}(l)\) were added? (d) How would the equilibrium be affected if the reaction vessel were warmed? (e) How would the equilibrium be affected if a catalyst were added?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free