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.
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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?
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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.
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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.
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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?
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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.)
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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?
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