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

K'c = 5.3 × 10⁻⁵

Step-by-step 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=([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=([NH_(3)]^(2) )/([N_(2)].[H_(2)]^(3) ) =([NH_(3)].[NH_(3)])/([N_(2)]^(1/2).[N_(2)]^(1/2). [H_(2)]^(3/2).[H_(2)]^(3/2)) =(([NH_(3)])/([N_(2)]^(1/2).[H_(2)]^(3/2)) )^(2) =(1)/(K'c^(2) ) \\K'c=\sqrt{(1)/(kc) } =\sqrt{(1)/(3.6 * 10^(8)  ) } =5.3 * 10^(-5)`