Question

After 1.60 mol of NH3 gas is placed in a 1600-cm^3 box at 25 degrees C, the box is heated to 500 K. At this temperature, the ammonia is partially decomposed to N2 and H2, and a pressure measurement gives 4.85 MPa. Find the number of moles of each component present at 500 K.

Answer 1

Given :

Amount of NH3 = 1.60 mol

Volume of box = 1600 cm³

T = 500 K

Pressure = 4.85 MPa.

Solution:

Lets find moles of NH3 by using ideal gas equation.

pV = nRT

Here p is pressure , V is volume, n number of moles R is gas constant value is 8.314 Pa m³ per K per mol T is temperature in K.

Rearrange the equation,

n = pV / RT

Before finding out moles we convert pressure to Pa and volume to m³

Pressure in atm = 4.85 Mpa X (1 E6 Pa / 1 Mpa )

                           = 4.85 E 6 Pa

Volume in m³   = 1600 cm³ x ( 1 m³ / 1 E6 cm³ )

                          = 1.6 E-3 m³

Calculation of moles :

n = pV / RT

   = (4.85 E 6 Pa X 1.6 E -3 m³ ) / ( 8.314 Pa m³ per K per mol X 500 K )

   = 1.867

Reaction of decomposition of NH3 and ICE

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

I           1.60                 0                     0

C          -2x                   +x           + 3x

E          (1.60 -2x)            x             3x

Total number of moles = ( 1.60 – 2x) + x + 3x = 1.867

Calculation of x

1.60 – 2x + x + 3x = 1.867

1.60 + 2x = 1.867

2x = 1.867 – 1.60

2x = 0.2667

x = 0.1334

Lets find equilibrium moles of each

Moles of NH3 = 1.60 – 2 x = 1.60 – 2 * 0.1334 = 1.333 mol NH3

Mol N2 = x = 0.1334 mol N2

Mol H2 = 3 x = 0.4 mol H2