Bomai places 7mol of Ammonia (NH3) in 14L flask. At equilibrium 0.2M Ammonium (NH4) was found in the flask. What is the equilibrium constant (Kc) for this reaction?
1. Write a balanced chemical reaction for the above reaction?
2. What is the equilibrium constant (Kc) for this reaction?
3. What are conjugate acids and base for the reaction in 1?
1. Balanced chemical reaction: NH3 (g) ⇌ NH4+ (aq) + OH- (aq)
2. To calculate the equilibrium constant (Kc), we need to use the concentrations of the reactants and products at equilibrium. Since the initial concentration of NH3 was 7mol in a 14L flask, the concentration of NH3 at equilibrium is 0.5 M (7 mol / 14 L = 0.5 M). The concentration of NH4+ at equilibrium is given as 0.2 M. The concentration of OH- at equilibrium can be calculated using the fact that NH3 reacts with water to form NH4+ and OH-. Since every molecule of NH3 forms one NH4+ and one OH-, the concentration of OH- at equilibrium is also 0.5 M.
The equilibrium constant (Kc) is calculated as [NH4+][OH-] / [NH3]. Therefore, Kc = (0.2)(0.5) / 0.5 = 0.2
3. The conjugate acid of NH3 is NH4+ and the conjugate base of NH4+ is NH3.