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

At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$.If 2.0 moles of \(\mathrm{NO}\) and 1.0 mole of \(\mathrm{Cl}_{2}\) are placed into a \(1.0-\mathrm{L}\) flask, calculate the equilibrium concentrations of all species.

Short Answer

Expert verified
The equilibrium concentrations of NOCl, NO, and Cl₂ are 0.074 mol/L, 1.926 mol/L, and 0.963 mol/L, respectively.

Step by step solution

01

Write the balanced reaction equation

The balanced reaction equation is given as: \[2\mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)\]
02

Set up an ICE table

Using the given information, we can set up an ICE table to keep track of the concentrations: | | NOCl | NO | Cl₂ | |---------|-------|------|-------| | Initial | 0 | 2.0 | 1.0 | | Change | +2x | -2x | -x | | Equil. | 2x | 2-2x | 1-x | In which x stands for the change in concentration.
03

Write the equilibrium expression

The equilibrium constant K expression is: \[K = \frac{[\mathrm{NO}]^2[\mathrm{Cl}_{2}]}{[\mathrm{NOCl}]^2}\]
04

Substitute equilibrium concentrations into the expression

Substitute equilibrium concentrations from the ICE table into the equilibrium constant expression and the given K value: \[1.6 \times 10^{-5} = \frac{(2- 2x)^2(1-x)}{(2x)^2}\]
05

Solve for x

Solve for x: \[x = 0.037\]
06

Calculate the equilibrium concentrations

Substitute the value of x back into the equilibrium row of the ICE table and calculate the equilibrium concentrations of all species: - Equilibrium concentration of NOCl = 2x = 2(0.037) = 0.074 mol/L - Equilibrium concentration of NO = 2 - 2x = 2 - 2(0.037) = 1.926 mol/L - Equilibrium concentration of Cl₂ = 1 - x = 1 - 0.037 = 0.963 mol/L So, the equilibrium concentrations of NOCl, NO, and Cl₂ are 0.074 mol/L, 1.926 mol/L, and 0.963 mol/L, respectively.

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, represented by the letter 'K', is a vital tool in understanding chemical equilibria.
It quantifies the ratio of the concentrations of products to reactants at equilibrium.
Each concentration is raised to the power of its respective coefficient in the balanced reaction equation.
For the given reaction,
  • the equilibrium constant expression is \[K = \frac{[\mathrm{NO}]^2[\mathrm{Cl}_{2}]}{[\mathrm{NOCl}]^2}\]
The K value indicates the extent of the reaction; a very small K, like in this problem with \(K = 1.6 \times 10^{-5}\), suggests that the reactants are predominant at equilibrium.
This helps predict the direction in which a reaction needs to go to reach equilibrium.
ICE Table
An ICE table is a systematic way to calculate changes in concentrations of species in a chemical reaction.
"ICE" stands for Initial, Change, and Equilibrium.
In this exercise, it helps track the concentration changes from initial to equilibrium states.
  • **Initial**: Enter the initial concentrations of reactants and products. If not provided, assume it’s zero for products that haven't formed yet.
  • **Change**: Represent the change in concentration during the reaction by adding or subtracting variables (like \(x\)). Use stoichiometry to relate these changes based on the balanced equation.
  • **Equilibrium**: Calculate the resulting concentrations by combining initial concentrations and changes. The ICE table, therefore, simplifies solving for equilibrium concentrations by visualizing the concentration shifts in each species throughout the reaction.
Chemical Equilibrium
Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the backward reaction.
At this point, the concentrations of all reactants and products remain constant over time.
This doesn’t mean that the concentrations are equal, but that their ratios remain constant as per the equilibrium constant. When a system reaches equilibrium:
  • The macroscopic properties, such as concentration, no longer change.
  • The microscopic processes continue to occur, meaning both forward and reverse reactions take place at equal rates.
Understanding this concept helps us grasp why reactions don't always proceed to completion and how conditions can affect the position of equilibrium.
Reaction Equation
The balanced reaction equation provides a complete and accurate representation of a chemical reaction.
It specifies the reactants and products involved, alongside their stoichiometry.
This is critical for writing the equilibrium expression and setting up ICE tables.
In this exercise, the balanced equation is:
  • \[2\mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)\]
This equation tells us:
  • The number of moles of each substance participating in the reaction, which guides us in calculating changes during the process using the ICE table.
  • The direction of the reaction, which can be predicted by comparing initial concentrations to those at equilibrium.
Understanding and correctly balancing the reaction equation is crucial for successful equilibrium calculations and predictions.

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

A sample of solid ammonium chloride was placed in an evacuated container and then heated so that it decomposed to ammonia gas and hydrogen chloride gas. After heating, the total pressure in the container was found to be 4.4 atm. Calculate \(K_{\mathrm{p}}\) at this temperature for the decomposition reaction $$\mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HCl}(g)$$,

Novelty devices for predicting rain contain cobalt(II) chloride and are based on the following equilibrium:$$\mathrm{CoCl}_{2}(s)+6 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CoCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}(s)$$ Purple Pink.What color will such an indicator be if rain is imminent?

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 -mole sample of naphthalene has a mass of 32.8 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)$$$K=4.29 \times 10^{-6}(\text {at } 298 \mathrm{K})$$.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 following exothermic reaction at equilibrium: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$. Predict how the following changes affect the number of moles of each component of the system after equilibrium is reestablished by completing the table below. Complete the table with the terms increase, decrease, or no change.

A sample of gaseous nitrosyl bromide (NOBr) was placed in a container fitted with a frictionless, massless piston, where it decomposed at \(25^{\circ} \mathrm{C}\) according to the following equation:$$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ The initial density of the system was recorded as \(4.495 \mathrm{g} / \mathrm{L}\) After equilibrium was reached, the density was noted to be \(4.086 \mathrm{g} / \mathrm{L}\) a. Determine the value of the equilibrium constant \(K\) for the reaction. b. If \(\mathrm{Ar}(g)\) is added to the system at equilibrium at constant temperature, what will happen to the equilibrium position? What happens to the value of \(K ?\) Explain each answer.

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