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
Ch.15 - Chemical Equilibrium
Chapter 15, Problem 88
A characteristic reaction of ethyl acetate is hydrolysis, the reverse of the reaction in Problem 15.87. Write the equilibrium equation for the hydrolysis of ethyl acetate, and use the data in Problem 15.87 to calculate Kc for the hydrolysis reaction.
Verified step by step guidance1
Step 1: Write the balanced chemical equation for the hydrolysis of ethyl acetate. The hydrolysis of ethyl acetate (CH3COOC2H5) in the presence of water (H2O) produces ethanol (C2H5OH) and acetic acid (CH3COOH). The equation is: CH3COOC2H5 + H2O \rightleftharpoons C2H5OH + CH3COOH.
Step 2: Identify the equilibrium expression for the reaction. The equilibrium constant expression (Kc) for the reaction is based on the concentrations of the products and reactants at equilibrium: Kc = \frac{[C2H5OH][CH3COOH]}{[CH3COOC2H5][H2O]}.
Step 3: Use the data from Problem 15.87 to determine the concentrations of the reactants and products at equilibrium. This involves substituting the given equilibrium concentrations into the Kc expression.
Step 4: Calculate the equilibrium constant (Kc) using the equilibrium concentrations. Substitute the equilibrium concentrations into the Kc expression and solve for Kc.
Step 5: Interpret the value of Kc. A large Kc value indicates that the reaction favors the formation of products, while a small Kc value suggests that the reactants are favored at equilibrium.
Related Practice
Open Question
A sample of HI 19.30 * 10^-3 mol^2 was placed in an empty 2.00-L container at 1000 K. After equilibrium was reached, the concentration of I2 was 6.29 * 10^-4 M. Calculate the value of Kc at 1000 K for the reaction H2(g) + I2(g) ⇌ 2 HI(g).
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
Open Question
Gaseous indium dihydride is formed from the elements at elevated temperature: ln(g) + H2(g)
ightarrow lnH2(g), Kp = 1.48 at 973 K. Partial pressures measured in a reaction vessel are: P_in = 0.0600 atm, P_H2 = 0.0350 atm, P_lnH2 = 0.0760 atm. (b) Determine the equilibrium partial pressures of all the gases.
Textbook Question
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.
265
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