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

The numerical value of equilibrium constant is 0.0560

Step-by-step explanation:

Initial Concentration:

`[NH^3]= (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 *(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 *(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 =`
`([NH^3]^2)/(N^2[H^2]^3)`

`K_c =`
`(1.96^2)/((1.52 )*(3.56)^3)`

`K_c =` 0.0560

Therefore, the numerical value of equilibrium constant is 0.0560