Question

Use the References to access impertant values if needed for this question. A student ran the following reaction in the laboratory at 700 K N2(g)+3H2(g) 2NH,(g) When she introduced 3.63x102 moles of N;(g) and 5.18x10 2 moles of H2(g) into a 1.00 liter container, she found the equilibrium concentration of NHs(g) to be 1.11x10 3 M. Calculate the equilibrium constant, Ke, Sshe obtained for this reaction. Ke= Submit Answer Retry Entire Group 7 more group attempts remaining

Answer 1

Solution :-

Balanced reaction equation

N2(g)+3H2(g) --- > 2NH3 (g)

Initial moles of N2 and H2 are 3.63*10^-2 mol and 5.18*10^-2 mol respectively

Equilibrium concentration of NH3=1.11*10^-3 M

Volume = 1 L

Calculating the initial concentration of N2 and H2

Molarity =moles/volume

Molarity of N2= 3.63*10^-2 mol /1.0 L =3.63*10^-2 M

Molarity of H2=5.18*10^-2 mol /1.0 L=5.18*10^-2 M

Making ICE table

     N2(g)    +         3H2(g)         --- >      2NH3 (g)

I 3.63*10^-2 M    5.18*10^-2 M              0

C    –x                            -3x                            +2x                         

E   3.63*10^-2 M-x   5.18*10^-2 M-3x    1.11*10^-3 M

Calculating the value of x

2x=1.11*10^-3 M

X=1.11*10^-3 M/2

X=5.55*10^-4 M

Calculating the equilibrium concentration of N2 and H2

[N2]eq=3.63*10^-2 M –x

             =3.63*10^-2 M -5.55*10^-4 M

             =3.57*10^-2 M

[H2]eq=5.18*10^-2 M -3x
             =5.18*10^-2 –(3*5.55*10^-4)

            =5.01*10^-2 M

Calculating the equilibrium constant

Kc=[NH3]^2/[N2][H2]^3

   =[1.11*10^-3]^2 / ([3.57*10^-2][5.01*10^-2]^3)

   =0.274

Therefore answer is Kc=0.274