Question

For the reaction N2(g) + 3 H2(g) = 2 NH3(g), Kp = 0.0489 at 256˚C. For parts a through d, assume that the reaction is at this temperature.

c) A third equilibrium mixture contains 0.0100 mol/L of N2 and 0.0300 mol/L of NH3.
What is the concentration of H2 in this mixture?

d) A fourth equilibrium mixture contains equal concentrations of all three chemicals.
What is the pressure of each substance in this mixture?

Please help me on c & d by 11/25 11pm thanks!

Answer 1

Step-by-step explanation:

`N_2(g)+3H_2(g)\rightleftharpoons 2NH_3(g)`

c) Concentration of
`H_2` in this mixture

`K_p=0.0489`

`\Delta n_g`:

=(gaseous mole on product side)-(gaseous mole on reactant side)=2-4=-2

R = 8.314 J/mol K

T = 256 °C = 256+273 = 529 K

`K_p=0.0489=K_c(RT)^(\Delta n_g)`

`K_c=940,524.93`

`K_c=940,524.93=([NH_3]^2)/([N_2][H_2]^3)=((0.0300 mol/L)^3)/(0.0100 mol/L* [H_2]^3)`

`[H_2]^3=10,450,277`

`[H_2]=218.62 mol/L`

The concentration of hydrogen gas is 218.62 mol/L

d)A fourth equilibrium mixture contains equal concentrations of all three chemicals

`[N_2]=[H_2]=[NH_3]=x mol/L`

`K_c=940,524.93=([NH_3]^2)/([N_2][H_2]^3)=(x^2)/(x* x^3)=(1)/(x^2)`

x =
`0.001031 mol/L=1.031* 10^(-3)mol/L`

According ideal gas equation:

PV=nRT

`P=(n)/(V)RT=\text{(total concentration of gases)}RT`

`=3* 1.031* 10^(-3)mol/L* 0.0821 L atm/mol K* 529K=0.1343 atm`

Total pressure,P = 0.1343 atm

Partial pressure of nitrogen gas
`p_(N_2)=P* \chi_(N_2)=0.1343*(1)/(3)=0.0447 atm`

Partial pressure of hydrogen gas
`p_(H_2)=P* \chi_(H_2)=0.1343*(1)/(3)=0.0447 atm`

Partial pressure of ammonia gas
`p_(NH_3)=P* \chi_(NH_3)=0.1343*(1)/(3)=0.0447 atm`