Question

A 1.000 L vessel is filled with 1.000 mole of N2,2.000 moles of H2, and 3.000 moles of NH3. When the reaction N2(g) + 3 H2(g)⇀↽2 NH3(g) comes to equilibrium, it is observed that the concentration of NH3is 2.12 moles/L. What is the numerical value of the equilibrium constant Kc?

Answer 1

Answer:

The numerical value of equilibrium constant is 0.0560

Explanation:

Initial Concentration:

$[NH^3]= \frac{3mol}{1l}$  = 3M

[$N_2$] = 1M

[$H_2$] = 2 M

$N_2 +3H_2 \rightleftarrows 2NH3$

at the end, [$NH_3$] = 1.96 M

Thus the change is 3 - 1.96  = 1.04 M

Thus 1.04 moles of  reacted

By stoichiometry, 1.04 moles  ($NH_3 \times\frac{1 mol N_2}{2 mol NH_3}$) = 0.52 mol created (in addition to 1 mol already in vessel)

By similar reasoning   = $1.04 \times\frac{3}{2 }$ 1.56 moles  created  

Final concentrations:

[$NH_3$] = 1.96 M

[$N_2$] = 0.52 + 1 = 1.52 M

[$H^2$] = 1.56 + 2  = 3.56 M

$K_c =$ $\frac{[NH^3]^2}{N^2[H^2]^3}$

$K_c =$  $\frac{1.96^2}{(1.52 )\times(3.56)^3}$

$K_c =$ 0.0560

Therefore, the numerical value of equilibrium constant is 0.0560