Question

Nitrogen gas rreacts with hydrogen gas to form ammonia. At 200 celcius iun a closed container, 1 atm of nitrogen gas is mixed with 2 atm of hydogen gas. At equilibnrium, the toal pressure is 2 atm. Calcualate the value of $K_p$?

Answer 1

Answer:

The value of Kp is 4

Explanation:

Step 1: Data given

The initial pressure of N2 = 1 atm

The initial pressure of  H2 = 2 atm

Temperature = 200 °C

At equilibnrium, the toal pressure is 2 atm

Step 2: The balanced equation

N2(g) + 3H2(g) → 2NH3(g)

Step 3: The initial pressures

N2 = 1.0 atm

H2 = 2.0 atm

NH3 = 0 atm

Step 4: The partial pressures at equilibrium

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

There will react X moles of N2,

There will react 3X moles of H2

There will be produced 2X moles NH3

The partial pressure of N2 at the equilibrium is (1.0 -X)atm

The partial pressure of H2 at the equilibrium is (2.0 - 3X)atm

The partial pressure of NH3 at the equilibrium is 2X

The total pressure at the equilibrium = 2.0 atm

The total pressure = pN2 + pH2 +pNH3 = 2.0 atm

2.0 atm = (1.0-X) + (2.0 - 3X) + 2X

2.0 =3.0 -2X

X = 0.50

The partial pressure of N2 at the equilibrium is (1.0 -0.50) = 0.50 atm

The partial pressure of H2 at the equilibrium is (2.0 - 3*0.5) = 0.50 atm

The partial pressure of NH3 at the equilibrium is 2*0.5 = 1.0 atm

Step 5: Calculate Kp

Kp = pNH3 / (pN2)*(pH2)

Kp = 1.0 / (0.5*0.5)

Kp = 4

The value of Kp is 4