Question

19) Consider the following reaction, equilibrium concentrations, and equilibrium constant at a particular temperature. Determine the equilibrium concentration of H2O(g). C2H4(8) + H2O(g) - C2H5OH(g) Kc = 9.0 x 103 (C2H4leq=0.015 M [C2H5OHleq=1.69 M A) 1.0 M B) 9.9 x 10-7 M C) 0.013 M D) 1.68 M E) 80.M 20) Consider the following reaction at equilibrium. What effect will increasing the volume of the reactic mixture have on the system? 20) 2 H2S(8) + 3 O2(g) + 2 H2O(g) +2 SO2(6) A) The reaction will shift to the left in the direction of reactants. B) The equilibrium constant will decrease. C) The equilibrium constant will increase. D) No effect will be observed. E) The reaction will shift to the right in the direction of products. 21) 21) Given the following balanced equation, determine the rate of reaction with respect to [02]. If the rate of formation of O2 is 6.94 x 10-1 M/s, what is the rate of the loss of O3? 203(g) – 302(B) A) 2.08 M/s B) 4.16 M/S 90.463 M/s D) 0.231 M/s E) 1.04 M/s

Answer 1

6 Given data ke=9.0 x103 [Ca Halee = 0.015M (CHE Orleq = 1-69 M Solution Ca H4 (9) + H₂O (8) CyHGOH (O) ke = [G H5OH] - [CH] [H₂O] [Heo] = [C₂H5OH [G HA]o ke [H0] = 169 0.0159.0 x103 [490] = 0.0125 [H20] 0.013 Hence option (c) is correct. 2H2S (7) + 30,(3) = 2H20 (9) + ascu () Number of moles of reactant - 5 Number of motep of products = 4. Since there are more moles of Reactants, an increase in Volume wild drift the equilibrium to the left in order to favor the reactants. Hence option (A) is correct.

8 2. Given data rate of formation of a = 6.94 x 10- 1 Solution 203 (8) 300 (8) The rate expression for above reaction is [0] = + d [0,7 I dt = { x6.94X10/min -d [0₂] = 0.463 M/A Since rate is always positive there fore - d[03] is taken as positive. Hence option (c) is correat.