ame: 13. Given the following reaction: Pbcro, (s)-Pb?\"(aq)+602 (aq) The equilib

Question

ame: 13. Given the following reaction: Pbcro, (s)-Pb?"(aq)+602 (aq) The equilibrium constant for the above reaction is given as 14. Consider the equilibrium reaction where carbon monoxide reacts with steam to produce hydrogen and carbon dioxide. The equilibrium position occurs far to the right. This means that once equilibrium has been established, a. very little product remains b. the reaction vessel would contain equal amounts of product and reactant c. very little reactant remains d. the reaction vessel contains high concentrations of reactant 15. The equilibrium system shown below was analyzed and the concentrations of H:(8). 1(g) and Hl(g), were found, in molL, to be 4.2, 3.8, 16respectively. The equilibrium constant must be which of the following? H(g)+14g)2HI(g)+ 65 k a. 0.10 b. 28 d. 2.3 e. 1.4 16. Q is used to denote a trial equilibrium constant. Which of the following statements is TRUE? a. b. c. d. e. IfQ K, it means the forward reaction IfQ-K, it means the reaction is at equilibrium. All of the above are true. None of the above are true. will proceed to form more products 17. If a chemical system at equilibrium is disturbed, the rates of the forward and reverse reactions are unequal as the system returns to equilibrium. This is known as a. the Law of Mass Action b. The Law of Chemical Equilibriunm c. the Haber-Bosch Process d. the Unequal Rate Principle e. Le Chatelier's Principle For the equilibrium system below, which of the following would result in an increase in the quantity of CO(g)? 18. 2H2(g) + CO(g) CH,0H(g) + 92 kJ i. decreasing temperature ii decreasing the volume of the container ii. adding some CH,OH iv. decreasing the pressure in the container v. removing some H d. ii., iv., and v. only e. all of i., ii, ii,iv. and v. a. i. and ii. only b. i., ii., and ii. only c. ii., ii. and v. only

Explanation / Answer

(13)

Reaction given:

PbCrO4(s) ---> Pb2+(aq) + CrO42-(aq)

The expression for Keq is:

Keq = Ksp = [Pb2+]*[CrO42-]

(14)

Since the equilibrium position occurs far to the right, this means that the conc of products is much high as compared to that of reactants.

So, very little reactant remains.

(15)

For the given reaction the expression for Keq is written as:

Keq = [HI]2/([H2]*[I2])

Putting values:

Keq = 1.62/(4.2*3.8) = 0.16

Hope this helps !

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