Part I A) The solubility of silver bromide can be increased by dissolving it in

Question

Part I
A) The solubility of silver bromide can be increased by dissolving it in a solution containing the thiosulfate anion.
AgBr(s) Ag+ (aq) + Br- (aq)    K1 = 7.7 x 10-13
Ag+(aq) + 2S2O32- (aq) Ag(S2O3)23- (aq)     K2 = 4.7 x 1013
What is the value of the equilibrium constant for the overall reaction?
AgBr(s) + 2S2O32-(aq) Ag(S2O3)23-(aq) + Br-(aq)
   a) 12.4x1013
b) 3.6x10-9
   c) 4.7x10-9
  d) 4.7x1013
   e) 36
B) Consider the following reaction at equilibrium. Which change will shift the equilibirum to the left?

   C3H8(g) + 5 O2(g)   3 CO2(g) + 4 H2O(l)          H° = -2220 kJ
   a) decreasing temperature
b) increasing temperature
     c) increasing volume
d) increasing the concentration of oxygen gas
C) Consider the following reaction at equilibrium. Which change will shift the equilibrium to the left?

   CO2(g) + 2 H2O(l) CH4(g) + 2 O2(g)       H° = +890 kJ
   a) decreasing temperature
   b) decreasing volume
   c) increasing amount of methane gas
   d) All of the above
   e) none of the above

Part II
A) Consider the following reaction at equilibrium. What effect will increasing the pressure of the reaction mixture have on the system?

   2 H2S(g) + 3 O2(g)   2 H2O(g) + 2 SO2(g)
a) The reaction will shift to the right in the direction of the products
   b) No effect will be observed
   c) The reaction will shift to the left in the direction of the reactants
   d) The equilibrium constant will decrease
   e) The equilibrium constant will increase

Explanation / Answer

Part I
A) The solubility of silver bromide can be increased by dissolving it in a solution containing the thiosulfate anion.
AgBr(s) Ag+ (aq) + Br- (aq)    K1 = 7.7 x 10-13
Ag+(aq) + 2S2O32- (aq)  Ag(S2O3)23- (aq)     K2 = 4.7 x 1013
What is the value of the equilibrium constant for the overall reaction?
AgBr(s) + 2S2O32-(aq) Ag(S2O3)23-(aq) + Br-(aq)

Solution :- equilibrium constant for the overall reaction = K1 * K2

                                                                                                 = 7.7*10^-13 * 4.7*10^13

                                                                                                 = 36

So the answer is option e

B) Consider the following reaction at equilibrium. Which change will shift the equilibirum to the left?

   C3H8(g) + 5 O2(g)   3 CO2(g) + 4 H2O(l)          H° = -2220 kJ

Solution :- reaction is exothermic means heat is given off by the reaction therefore increasing the temperature will cause the reaction to shift to left side.

So the answer is option b

C) Consider the following reaction at equilibrium. Which change will shift the equilibrium to the left?

   CO2(g) + 2 H2O(l) CH4(g) + 2 O2(g)       H° = +890 kJ

Solution :- Reaction is endothermic therefore heat is needed to proceed the reaction in the formward direction therefore decreasing the temperature will cause the equilibrium to shift to left side

So the answer is option a

Part II
A) Consider the following reaction at equilibrium. What effect will increasing the pressure of the reaction mixture have on the system?

   2 H2S(g) + 3 O2(g)   2 H2O(g) + 2 SO2(g)

Solution :- there are total 5 gas moles on reactant side and 4 gas moles of product side

Therefore increasing the pressure will shift the equilibrium to right side

So the answer is option a

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