Question

Given the following reaction and its equilibrium constant at a certain temperature, N2(g) + 3H2(g) ⇌ 2NH3(g) Kc= 3.6x10^8

Calculate the numerical value of the equilibrium constant for the following reaction at the same temperature.

NH3(g) ⇌ 1/2 N2(g) + 3/2 H2 (g) Please show work.

Answer 1

Answer:

K'c = 5.3 × 10⁻⁵

Explanation:

Let's consider the following reaction.

NH₃(g) ⇌ 1/2 N₂(g) + 3/2 H₂(g)

The equilibrium constant K'c is:

$K'c=\frac{[N_{2}]^{1/2}.[H_{2}]^{3/2}  }{[NH_{3}]}$

Now, let's consider this other reaction with Kc = 3.6 × 10⁸.

N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g)

$Kc=\frac{[NH_{3}]^{2} }{[N_{2}].[H_{2}]^{3} } =\frac{[NH_{3}].[NH_{3}]}{[N_{2}]^{1/2}.[N_{2}]^{1/2}. [H_{2}]^{3/2}.[H_{2}]^{3/2}} =(\frac{[NH_{3}]}{[N_{2}]^{1/2}.[H_{2}]^{3/2}} )^{2} =\frac{1}{K'c^{2} } \\K'c=\sqrt{\frac{1}{kc} } =\sqrt{\frac{1}{3.6 \times 10^{8}  } } =5.3 \times 10^{-5}$