Question

Ammonia decomposes to form nitrogen and hydrogen, like this:

2NH3(g) → N2(g) + 3H2(g)
Also, a chemist finds that at a certain temperature the equilibrium mixture of ammonia, nitrogen, and hydrogen has the following composition:
compound pressure at equilibrium
NH3 8.11atm
N2 84.5atm
H2 39.9atm
Calculate the value of the equilibrium constant Kp for this reaction. Round your answer to 2 significant digits.
=Kp

Answer 1

The equilibrium constant Kp for the decomposition of ammonia into nitrogen and hydrogen at the given pressures is approximately 5.4 x 10^7 when rounded to two significant digits.

Step-by-step explanation:

To calculate the value of the equilibrium constant Kp for the decomposition of ammonia into nitrogen and hydrogen, we can use the given pressures at equilibrium for NH3, N2, and H2. The balanced chemical equation is 2NH3(g) → N2(g) + 3H2(g). According to this equation, the Kp expression is:

Kp = (PN2)(PH2)3 / (PNH3)2

Plugging in the values:

Kp = (84.5 atm)(39.9 atm) / (8.11 atm)2

Calculating gives us:

Kp = (84.5 x 39.93) / 8.112

Kp = (84.5 x 63605.199) / 65.7721

Kp ≈ 5.38 x 107

Therefore, Kp for this reaction at the given temperature, rounded to two significant digits, is 5.4 x 107.