Question

use the references to access important values if needed for this question.
the equilibrium constant, ( k_{p} ) for the following reaction is ( 2.20\times10^{4} ) at 723 k.
( 2nh_{3}(g)
ightleftharpoons n_{2}(g)+3h_{2}(g) )
if an equilibrium mixture of the three gases in a 16.3 l container at 723 k contains ( nh_{3} ) at a pressure of 0.579 atm and ( n_{2} ) at a pressure of 0.757 atm, the equilibrium partial pressure of ( h_{2} ) is atm.
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Explanation:

Step1: Write the expression for \(K_p\)

For the reaction \(2NH_3(g)
ightleftharpoons N_2(g)+3H_2(g)\), the equilibrium constant expression \(K_p=\frac{P_{N_2}\times P_{H_2}^3}{P_{NH_3}^2}\)

Step2: Substitute the known values into the \(K_p\) expression

We know that \(K_p = 2.20\times10^{4}\), \(P_{NH_3}=0.579\ atm\), and \(P_{N_2}=0.757\ atm\).
Substituting these values into \(K_p=\frac{P_{N_2}\times P_{H_2}^3}{P_{NH_3}^2}\), we get \(2.20\times 10^{4}=\frac{0.757\times P_{H_2}^3}{(0.579)^2}\)

Step3: Solve for \(P_{H_2}^3\)

First, calculate \((0.579)^2 = 0.335241\). Then, rewrite the equation as \(P_{H_2}^3=\frac{2.20\times 10^{4}\times0.335241}{0.757}\)
\(P_{H_2}^3=\frac{7375.302}{0.757}\approx9742.8\)

Step4: Solve for \(P_{H_2}\)

Take the cube - root of both sides: \(P_{H_2}=\sqrt[3]{9742.8}\approx21.3\)

Answer:

\(21.3\)