Question

ammonia will decompose into nitrogen and hydrogen at high temperature. an industrial chemist studying this reaction fills a 500. ml flask with 3.6 atm of ammonia gas, and when the mixture has come to equilibrium measures the partial pressure of hydrogen gas to be 3.8 atm. calculate the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture. round your answer to 2 significant digits.

Explanation:

Step1: Write the balanced chemical equation

The decomposition of ammonia is \(2NH_3(g)
ightleftharpoons N_2(g)+3H_2(g)\)

Step2: Set up an ICE table (Initial - Change - Equilibrium)

	\(P_{NH_3}\)	\(P_{N_2}\)	\(P_{H_2}\)
Change	\(- 2x\)	\(+x\)	\(+3x\)
Equilibrium	\(3.6 - 2x\)	\(x\)	\(3x\)

We know that at equilibrium \(P_{H_2}=3.8\ atm\). Since \(P_{H_2} = 3x\), then \(3x=3.8\), so \(x=\frac{3.8}{3}\approx1.27\ atm\)

Step3: Calculate the equilibrium partial pressure of \(NH_3\)

\(P_{NH_3}=3.6-2x\). Substitute \(x = \frac{3.8}{3}\) into the formula:
\(P_{NH_3}=3.6-2\times\frac{3.8}{3}=3.6-\frac{7.6}{3}=\frac{10.8 - 7.6}{3}=\frac{3.2}{3}\approx1.07\ atm\)
\(P_{N_2}=x\approx1.27\ atm\)

Step4: Calculate the pressure equilibrium constant \(K_p\)

The formula for \(K_p\) is \(K_p=\frac{P_{N_2}\times P_{H_2}^3}{P_{NH_3}^2}\)
Substitute \(P_{NH_3}\approx1.07\ atm\), \(P_{N_2}\approx1.27\ atm\) and \(P_{H_2} = 3.8\ atm\) into the formula:
\(K_p=\frac{1.27\times(3.8)^3}{(1.07)^2}\)
First, calculate \((3.8)^3=3.8\times3.8\times3.8 = 54.872\)
Then \(1.27\times54.872 = 69.68744\)
And \((1.07)^2=1.1449\)
\(K_p=\frac{69.68744}{1.1449}\approx61\)

Answer:

\(61\)