7 moles of ammonia is placed in 14L flask.At equilibrium 0.2M ammonium was found in the flask. What is the equilibrium constant for this reaction.
1. Write the balance chemical equation for the above reaction
2. What is the equilibrium constant for this reaction
3. What are the conjugate acid and base for the reaction
1. The balanced chemical equation for the reaction is:
NH3 (g) ⇌ NH4+ (aq) + OH- (aq)
2. The equilibrium constant (K) for the reaction is given by:
K = [NH4+] [OH-] / [NH3]
Given that the concentration of NH4+ at equilibrium is 0.2 M and the initial concentration of NH3 was 7 moles in a 14 L flask (0.5 M), we can calculate the concentration of OH- as 0.2 M using the equation for the reaction. Substituting these values into the equilibrium constant equation, we get:
K = (0.2)(0.2) / 0.5
K = 0.08 / 0.5
K = 0.16
Therefore, the equilibrium constant for this reaction is 0.16.
3. The conjugate acid-base pair for the reaction is:
NH3 (weak base) and NH4+ (conjugate acid)