contestada

Consider the gas-phase reaction:
N2(g) + 3H2(g) ⇆ 2NH3 (g)
, for which Kp = 43 at 400 K.
If the mixture is analyzed and found to contain 0.18 bar of N2, 0.36 bar of H2 and 0.59 bar of NH3, describe the situation:

a) Q > K and more products will be made to reach equilibrium.
b) Q < K and more reactants will be made to reach equilibrium.
c) Q < K and more products will be made to reach equilibrium.
d) Q > K and more reactants will be made to reach equilibrium.
e) Within 1 decimal place, Q = K and the reaction is at equilibrium.

Respuesta :

Answer:

Q < K and more products will be made to reach equilibrium (option C)

Explanation:

Step 1: Data given

Kp = 43 at 400 K

pressure of N2 = 0.18 bar

Pressure of H2 = 0.36 bar

Pressure of NH3 = 0.59 bar

Step 2: The balanced equation

N2(g) + 3 H2(g) <=> 2 NH3 (g),  

Step 3: Calculate Q

Kp = 43  =  [NH3]²  /  [N2] [H2]³

Q =  [NH3]²  /  [N2] [H2]³

Q =  [0.59]²   /  [0.18] [0.36]³

Q = 0.3481 / 0.00839808

Q = 41.45

Q < Kp

When Q<Kp, there are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right.

Option C is correct (Q < K and more products will be made to reach equilibrium)

Otras preguntas