Question

a chemical engineer is studying the following reaction:

n₂(g)+3h₂(g)→2nh₃(g)

at the temperature the engineer picks, the equilibrium constant k, for this reaction is 0.00099.

the engineer charges (\fills\) three reaction vessels with nitrogen and hydrogen, and lets the reaction begin. she then measures the composition of the mixture inside each vessel from time to time. her first set of measurements are shown in the table below.

predict the changes in the compositions the engineer should expect next time she measures the compositions.

Explanation:

Step1: Calculate the reaction quotient \(Q_p\) for each vessel

The formula for \(Q_p\) for the reaction \(N_2(g)+3H_2(g)
ightleftharpoons 2NH_3(g)\) is \(Q_p=\frac{P_{NH_3}^2}{P_{N_2}\times P_{H_2}^3}\)

• For vessel A:

\(P_{N_2} = 35.98\ atm\), \(P_{H_2}=39.05\ atm\), \(P_{NH_3}=43.86\ atm\)
\(Q_{pA}=\frac{(43.86)^2}{35.98\times(39.05)^3}\)
\(Q_{pA}=\frac{1923.7}{35.98\times59597.7}\)
\(Q_{pA}=\frac{1923.7}{2.145\times10^6}\approx0.000897\)

• For vessel B:

\(P_{N_2} = 35.70\ atm\), \(P_{H_2}=38.20\ atm\), \(P_{NH_3}=44.43\ atm\)
\(Q_{pB}=\frac{(44.43)^2}{35.70\times(38.20)^3}\)
\(Q_{pB}=\frac{1974.0}{35.70\times55741.7}\)
\(Q_{pB}=\frac{1974.0}{1.99\times10^6}\approx0.000992\)

• For vessel C:

\(P_{N_2} = 40.98\ atm\), \(P_{H_2}=43.85\ atm\), \(P_{NH_3}=58.54\ atm\)
\(Q_{pC}=\frac{(58.54)^2}{40.98\times(43.85)^3}\)
\(Q_{pC}=\frac{3427.9}{40.98\times84077.7}\)
\(Q_{pC}=\frac{3427.9}{3.44\times10^6}\approx0.0010\)

Step2: Compare \(Q_p\) with \(K_p = 0.00099\)

• For vessel A:

Since \(Q_{pA}(0.000897)<K_p(0.00099)\), the reaction will shift to the right (towards the formation of \(NH_3\)). So, \(P_{N_2}\) will \(\downarrow\) (decrease), \(P_{H_2}\) will \(\downarrow\) (decrease), and \(P_{NH_3}\) will \(\uparrow\) (increase)

• For vessel B:

Since \(Q_{pB}(0.000992)\approx K_p(0.00099)\) (slightly greater), the reaction will shift to the left (towards the formation of \(N_2\) and \(H_2\)). So, \(P_{N_2}\) will \(\uparrow\) (increase), \(P_{H_2}\) will \(\uparrow\) (increase), and \(P_{NH_3}\) will \(\downarrow\) (decrease)

• For vessel C:

Since \(Q_{pC}(0.0010)>K_p(0.00099)\), the reaction will shift to the left (towards the formation of \(N_2\) and \(H_2\)). So, \(P_{N_2}\) will \(\uparrow\) (increase), \(P_{H_2}\) will \(\uparrow\) (increase), and \(P_{NH_3}\) will \(\downarrow\) (decrease)

Answer:

• Vessel A:
• \(N_2\): \(\downarrow\) (decrease)
• \(H_2\): \(\downarrow\) (decrease)
• \(NH_3\): \(\uparrow\) (increase)
• Vessel B:
• \(N_2\): \(\uparrow\) (increase)
• \(H_2\): \(\uparrow\) (increase)
• \(NH_3\): \(\downarrow\) (decrease)
• Vessel C:
• \(N_2\): \(\uparrow\) (increase)
• \(H_2\): \(\uparrow\) (increase)
• \(NH_3\): \(\downarrow\) (decrease)