Question

N2(g) + 3 H2(g) ⇌ 2 NH3(g) KP = 6.78 x 105 at 298 K (determined using atm)

A 7.5 x 101 L container being held at 298 K is charged with the three gases present in the above equation. Once finished, the initial partial pressure of N2 was 0.59 atm, the initial partial pressure of H2 was 0.45 atm, and the initial partial pressure of NH3 was 0.11 atm. The gas mixture was then allowed to reach equilibrium. Use this information to determine the partial pressures for all three gasses after equilibrium is reached.

Answer 1

Attached below is the handwritten solution to the above-mentioned problem.

N2 + 3 H2 = 2NH, ; Kp = 6.78x10 Step 1: Determine the reaction quotient) - Q= PNike (0.11² - 0.255< ke PNX PH (0.59)*(0:45) equilibrium shifts to e products Step 2 Make a chart describing relationships in change. + 3H, = 2NH. 0.45 0.11 N. Minitial / 0.59 change -x equil. (0.59-20) +2x -31 10.45-32) 10.11+236) Step 3 expression : Plug T into equilibrium and solve for a 6.78X10% - (0:11+2x)? (0.59-x)(0.45 -3x) Now we need to solve for ac. This is the tricky part. We can use approximations to get a nearby value of x. Or we can solve it in a numerical solver.

. We know that value of x will never be greater than either of the reactants. is. x<0.59 and 3 x < 0.45= 2<.0.15 => 1C 20.15 is the limiting case at nc = 0.15 kp becomes infinity but when nc is less than 0.15 our wndition holds We also notice that for x = 0.15, the maxima of x possible, the numerator becomes (0.11 + 2x0.15)² = 0.168 } maxima (0:11 + 2x0)= 0.012 į minima Therefore we get a rough idea that denominator gives the power to the Kp. it. . if x=0.14 (a value close to 0.15) we get 62 ao 8.215 x 104 → undicating 2 € (0.14,0.15)

- - On good further enough guessing we see that a = 0.146 is approximation for a. a 1. Equilibrium come are: PN = 0.59-0.146 = 0.444 calm PH₂ = 0.45-3x0.146 = 0.400 atm PNH. = 0.11+ 2x0.146 2 0.139 atm 22 L

On Solving we get the partial pressure of the gases at equilibrium as: (Approximately)

PN2 = 0.444atm

PH2 = 0.406atm

PNH3 = 0.139atm