Question

Consider the following equilibrium: Ag (aq) 2 NH3(aq) Ag(NH)21'(aq) AH15 k How does each of the following situations affect the position of the equilibrium? 1. Dissolving AgNO in the solution. (Remember that metal nitrates are fully soluble in water 2. 3. 4. and will dissociate upon dissolving: MfNOJy(s)->xMn"(aq) + yNO3-(aq).) Bubbling NH, (g) into the solution. (Remember that NIH, gas is soluble in water) Heating the solution. Dissolving NaCl in the solution, given the following information AgClo Ag(a) Cl(aq) K-1.8x 10-10

Answer 1

Ag⁺ (aq) + 2 NH₃(aq) $\rightleftharpoons$ [Ag(NH₃)₂]⁺ (aq)   $\Delta$ H=-15kJ

1. According to Le-Chatelier's principle when a reaction condition is changed the reaction will proceed in a direction which will try to maintain the equilibrium condition.

When AgNO₃ is added to the mixture, it dissociated into Ag⁺ and NO₃^-. Thus, the Ag+ concentration starts increasing. Thus, the reaction will proceed in forward reaction to maintain the equilibrium.

2. Similarly, when NH₃ gas is bubbled into the mixture, the reactant side concentration increases and hence more product will be formed to maintain the equilibrium.

3.The reaction given is an exothermic reaction i.e. heat is evolved when the reactant bonds are broken and new bonds for the product is formed. This suggests that more energy is required for the bond formation. So, when the reaction is heated (more energy provided) more products can be formed and thus to stop that and maintain equilibrium the reaction will shift in the backward direction.

4. NaCl when dissolved will dissociate to Na⁺ and Cl^-. The Cl- ions can combine with Ag+ to give AgCl precipitate. The k_eq value for AgCl given is very less which suggests that the dissociation of AgCl is less probable. Thus, upon adding NaCl to the reaction mixture, Ag+ concentration for the reaction with NH₃ is decreasing and hence product cannot be formed easily. Thus, the reaction shifts to backward side to maintain equilibrium.