Question

When 2.00 moles of ammonia are introduced into a 1.00 liter flask, it reacts as follows 2NH3(g)->3H2(g)+N2(g). At equilibrium 1.25 moles of ammonia remain. Calculate Kc

Answer 1

Kc = 0.101 or 1.01*10^-1

Step-by-step explanation:

Set up ICE table
2NH3 -> 3H2 + N2
I: 2 moles, 0 moles, 0 moles

C: -x moles +y moles +z moles

E: 1.25 moles 0+y moles 0+z moles

We know that x is 0.75 moles, since 2-0.75 is 1.25

for every mole of NH3 consumed, 1.5 moles of H2 are produced, and 0.5 moles of N2 are produced. If 0.75 moles of NH3 are consumed, then 1.5 * 0.75 moles of H2 are produced, and 1.5*0.5 moles of N2 are produced

so, y equals 1.5 * 0.75, which is 3/2 * 3/4, or 3/8, or 0.375 moles of H2 produced

z equals 1.5 * 0.5, which is 0.75 moles of N2 produced

Kc equals
`([N_2][H_2]^3)/([NH_3]^2)`

N2 = 0.75, N2 = 0.375, and NH3 = 1.25
So, Kc equals
(0.375 * 0.75^3)/(1.25^2), which equals 0.10125

Round to 3 sig figs to get 0.101