Question

A flask is charged with 1.550 atm of N2O4(g)and 1.00 atm NO2(g) at 25 ∘C, and the following equilibrium is achieved:
N2O4(g)⇌2NO2
After equilibrium is reached, the partial pressure of NO2 is 0.519 atm .

1.What is the equilibrium partial pressure of N2O4?Express your answer with the appropriate units.

2. Calculate the value of Kp for the reaction.

3.Calculate the value of Kc for the reaction.

Answer 1

1)

P_N2O4 = 1.791

Explanation

N2O4(g) <--------> 2NO2(g)

K_p = (P_NO2)²/P_N2O4

Initial partial pressure

P_N2O4 = 1.550

P_NO2 = 1.00

Change in partial pressure

P_N2O4 = - x

P_NO2 = + 2x

Equilibrium partial pressure

P_N2O4 = 1.550 - x

P_NO2 = 1.00 + 2x

at equilibrium ,

P_NO2= 0.519

1.00 + 2x = 0.519

2x = -0.481

x = -0.2405

P_N2O4 = 1.550 - (-0.2405) = 1.791

2)

K_p = 0.150

Explanation

K_p = (P_NO2)²/P_N2O4

K_p = (0.519)²/1.791

K_p = 0.150

3)

K_c = 0.00613

Explanation

K_p = K_c(RT)^∆ng

K_c = K_p/(RT)^∆ng

​​​​​where,

R = gas constant , 0.082057(L atm/mol K)

T = Temperature , 25℃ = 298.15K

∆n_g = number of moles of gaseous products - number of moles of gaseous reactants

= 2 - 1

= 0

K_c = 0.150atm/ ( 0.082057(L atm/mol K) × 298.15K)¹

K_c = 0.00613