Skip to main content
Ch.15 - Chemical Equilibrium
Chapter 15, Problem 90

The following reaction, which has Kc = 0.145 at 298 K, takes place in carbon tetrachloride solution: 2 BrCl1soln2 ∆ Br21soln2 + Cl21soln2 A measurement of the concentrations shows 3BrCl4 = 0.050 M, 3Br24 = 0.035 M, and 3Cl24 = 0.030 M. (b) Determine the equilibrium concentrations of BrCl, Br1, and Cl2.

Verified Solution

Video duration:
4m
This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Equilibrium Constant (Kc)

The equilibrium constant (Kc) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. It is calculated using the formula Kc = [products] / [reactants], where the concentrations are raised to the power of their coefficients in the balanced chemical equation. A Kc value less than 1 indicates that reactants are favored, while a value greater than 1 suggests that products are favored.
Recommended video:
Guided course
03:20
Equilibrium Constant Expressions

ICE Table (Initial, Change, Equilibrium)

An ICE table is a tool used to organize the initial concentrations, the changes in concentrations as the reaction proceeds, and the equilibrium concentrations of reactants and products. By setting up an ICE table, one can systematically determine how the concentrations of each species change from the initial state to equilibrium, allowing for the calculation of unknown concentrations using the equilibrium constant.
Recommended video:
Guided course
01:14
ICE Charts and Equilibrium Amount

Stoichiometry of the Reaction

Stoichiometry refers to the quantitative relationship between the reactants and products in a chemical reaction, as dictated by the balanced chemical equation. In this case, the stoichiometry indicates that two moles of BrCl produce one mole of Br2 and one mole of Cl2. Understanding the stoichiometric ratios is essential for calculating the changes in concentrations of each species during the reaction and for applying the equilibrium constant correctly.
Recommended video:
Guided course
01:16
Stoichiometry Concept
Related Practice
Textbook Question
The value of Kc for the reaction 3 O21g2 ∆ 2 O31g2 is 1.7 * 10-56 at 25°C. Do you expect pure air at 25 °C to contain much O3 (ozone) when O2 and O3 are in equilib- rium? If the equilibrium concentration of O2 in air at 25 °C is 8 * 10-3 M, what is the equilibrium concentration of O3?
601
views
Textbook Question
Calculate the equilibrium concentrations at 25 °C for the reaction in Problem 15.84 if the initial concentrations are 3N2O44 = 0.0200 M and 3NO24 = 0.0300 M.
570
views
Textbook Question
The industrial solvent ethyl acetate is produced by the reac-tion of acetic acid with ethanol: CH3CO2H1soln2 + CH3CH2OH1soln2 ∆ CH3CO2CH2CH31soln2 + H2O1soln2 Ethyl acetate (b) A solution prepared by mixing 1.00 mol of acetic acid and 1.00 mol of ethanol contains 0.65 mol of ethyl ace- tate at equilibrium. Calculate the value of Kc. Explain why you can calculate K without knowing the volume of the solution.
644
views
Textbook Question
An equilibrium mixture of N2, H2, and NH3 at 700 K con- tains 0.036 M N2 and 0.15 M H2. At this temperature, Kc for the reaction N21g2 + 3 H21g2 ∆ 2 NH31g2 is 0.29. What is the concentration of NH3?
813
views
Textbook Question
Recalculate the equilibrium concentrations in Problem 15.93 if the initial concentrations are 2.24 M N2 and 0.56 M O2. (This N2>O2 concentration ratio is the ratio found in air.)
296
views
Textbook Question
The value of Kc for the reaction of acetic acid with ethanol is 3.4 at 25°C: CH3CO2H1soln2 + CH3CH2OH1soln2 ∆ Acetic acid Ethanol CH3CO2CH2CH31soln2 + H2O1soln2 Kc = 3.4 (a) How many moles of ethyl acetate are present in an equi- librium mixture that contains 4.0 mol of acetic acid, 6.0 mol of ethanol, and 12.0 mol of water at 25 °C?
1319
views