Question

5.0 mole of ammonia were introduced into a 5.0 L reaction chamber in which it is partially decomposed at high temperature. At equilibrium at a particular temperature,80% of the ammonia had reacted. calculate Kc?

Answer 1

The decomposition reaction is below:

2NH₃ --> N₂ + 3H₂

The initial concentration of NH₃ = $\frac{5.0\ mol}{5.0\ L}=1.0\ M$

Knowing that 80% of NH₃ had reacted, the final concentration of NH₃ is 0.20 M. Based on this, an ICE table can be constructed:

Therefore,

$K_c=\frac{[N_2][H_2]^3}{[NH_3]^2}=\frac{0.40\times(1.2)^3}{(0.20)^2}\approx17$

Answer: 17