Consider the simple one-step reaction: A(g) = B(g) Since the reaction occurs in a single step, the forward reaction has a rate of kjoe[A] and the reverse reaction has a rate of kgev[B]. What happens to the rate of the forward reaction when we in- crease the concentration of A? How does this explain the reason behind Le Châtelier's principle?

Consider the simple one-step reaction: A(g) = B(g) Since the reaction occurs in a single step, the forward reaction has a rate of kjoe[A] and the reverse reaction has a rate of kgev[B]. What happens to the rate of the forward reaction when we in- crease the concentration of A? How does this explain the reason behind Le Châtelier's principle?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

According to LeChatelier's principle, when the reaction quotient is less than the equilibrium constant of a reaction, the reaction: A must shift toward products to reach equilibrium. B must shift toward reactants to reach equilibrium. (c) results in an increase in standard enthalpy of formation of reactants. results in an increase in standard enthalpy of formation of products.
6. Write the equilibrium constant expressions for the following reactions. a. 2Zns (s) + 302 (g) = 2Zn0(s) + 2S02 (g) b. C (s) + CO2 (g) = 2c0 (g) c. CH;CO,H (aq) = CH;COF (aq) + H* (aq)
1. Write the equilibrium constant expressions Ke for the following reactions. a) CHa(g) + H20 (g) →CO(g) + 3 H2 (g) b) 2 N2H4 (I) + 2 NO2 (g) → 3 N2 (g) + 4 H2O (I) c) NHẠNO3 (s) 2 H20 (g) + N20 (g) d) 4 HCl (aq) + MnO2 (s) = MnCl2 (aq) + 2 H2O (1) + Cl2 (g) e) Ca3(PO4)2 (s) = 3 Ca²* (aq) + 2 PO,³- (aq)
A student working in the laboratory prepared the following reactants: 5 mL of 0.008M Cd2+(aq) 10 mL of 0.006M SCN-(aq) 10 mL of 0.9M HNO3(aq) These reagents were mixed and allowed to stand for 10 minutes. The concentration of Cd(SCN)+ in the resulting equilibrium mixture is found to be 5 x 10−4M. Calculate the initial concentration of Cd2+(aq). Calculate the initial concentration of SCN-(aq). Calculate the equilibrium concentration of Cd2+(aq).
25- 2 H2S(g) → 2 H2(g) + S2(g) When heated, hydrogen sulfide gas decomposes according to the equation above. A 3.40 g sample of H2S(g) is introduced into an evacuated rigid 1.25 L container. The sealed container is heated to 483 K, and 3.72x10-2 mol of S2(g) is present at equilibrium. (a) Write the expression for the equilibrium constant, Ke, for the decomposition reaction represented above. (b) Calculate the equilibrium concentration, in mol·L-1, of the following gases in the container at 483 K. (i) H2(g) (i) Н2S(g) (c) Calculate the value of the equilibrium constant, Ko, for the decomposition reaction at 483 K. (d) Calculate the partial pressure of S2(3) in the container at equilibrium at 483 K. (e) For the reaction H2(g) + S2(2) –→ H2S(g) at 483 K, calculate the value of the equilibrium constant, K.
Suppose the following exothermic reaction UO2(s)+4HF(g) UFA(g) + 2H20(g) has already reached equilibrium. Predict the effect that each of the following changes will have on the equilibrium position. State whether the equilibrium will shift to the reactants, will shift to the products, or will not be affected. Explain your answer using Le Chatelier's Principle. a. The system was cooled. b. The reaction vessel was quadrupled. C. Uranium (IV) oxide was added to the reaction. d. The partial pressure of Uranium (IV) fluoride was decreased.
For the reversible reaction: N 2 (g)+ O 2 (g) = 2 NO (g) AH= 181 kJ (a) What will happen to the concentrations of NO at equilibrium if O, is added? (b) What will happen to the concentrations of NO at equilibrium if NO is removed? (c) What will happen to the concentrations of NO at equilibrium if the volume of the reaction vessel is increased? (d) What will happen to the concentrations of NO at equilibrium if the temperature of the system is decreased? (e) What will happen to the concentrations of N,, 0,, and NO at equilibrium if a catalyst is added?
Hydrofluoric acid and water react to form fluoride anion and hydronium cation, like this: + HF(aq)+H,O(1)→→F (aq)+H;O"(aq) At a certain temperature, a chemist finds that a 9.1 L reaction vessel containing an aqueous solution of hydrofluoric acid, water, fluoride anion, and hydronium cation at equilibrium has the following composition: compound amount HF 1.65 g H,O 745. g F 0.150 g H;O* 0.136 g Calculate the value of the equilibrium constant K, for this reaction. Round your answer to 2 significant digits. K = x10 ?
The equilibrium constant for the reaction 2 N2(g) + 5 O2(g) ⇆ 2 N2O5(g) was measured at two temperatures: Temperature, K Kp 343 2.5x10−8 944 7.6x10−6 What is ΔHo for the reaction (in kJ/mol)?
For the following reaction write out the equilibrium law
4. Sketch a graphical representation of the following events for the equilibrium below. PCIse 5 PCe + Cl ΔΗ = + 88 kJ/mol 15(g) 3(g) 2(g) a) PCI, is introduced into a flask at t=0 to a concentration of 2 mol/L. b) The flask is heated gently until no further changes in concentration of PCl, occur (5 minutes). At this time, the concentration of PCl(g) is 1.25 mol/L. c) At t=10 min an additional 0.25 mol/L of chlorine gas is injected into the flask. d) At t=15 min, no further changes in concentration were observed. Indicate the places on the graph where the system is at equilibrium. f) Use the information provided to calculate K, the equilibrium constant of this reactions. Conc'n (mol/L) Time (min)
Consider the reaction: 4 A (g) + 3 B (g) 2 C (g) + 5 E (g). The reaction is initiated with 0.60 M in A, 0.45 M in B and 0.50 M in E. At equilibrium, the value of [A] = 0.36 M. What is the equilibrium value of [E]?