Question

at a certain temperature, the equilibrium constant ( k ) for the following reaction is ( 0.0093 ):

( mathrm{n}_{2}(mathrm{~g})+mathrm{o}_{2}(mathrm{~g})
ightleftharpoons 2 mathrm{no}(mathrm{g}) )

use this information to complete the following table.

suppose a ( 21. mathrm{~l} ) reaction vessel is filled with ( 1.8 mathrm{~mol} ) of ( mathrm{no} ). what can you say about the composition of the mixture in the vessel at equilibrium?

( \bigcirc ) there will be very little ( mathrm{n}_{2} ) and ( mathrm{o}_{2} ).<br>( \bigcirc ) there will be very little no.<br>( \bigcirc ) neither of the above is true.

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what is the equilibrium constant for the following reaction?<br>round your answer to 2 significant digits.<br>( 2 mathrm{no}(mathrm{g})

ightleftharpoons mathrm{n}_{2}(mathrm{~g})+mathrm{o}_{2}(mathrm{~g}) ) | ( k=square ) |<br>| what is the equilibrium constant for the following reaction?<br>round your answer to 2 significant digits.<br>( 3 mathrm{~n}_{2}(mathrm{~g})+3 mathrm{o}_{2}(mathrm{~g})
ightleftharpoons 6 mathrm{no}(mathrm{g}) ) | ( k=square ) |

Explanation:

Step1: Analyze the first question

Given \(K = 0.0093\) for \(N_{2}(g)+O_{2}(g)
ightleftharpoons 2NO(g)\). The reaction quotient \(Q\) for the reverse process (starting with \(NO\)) will be very large initially (\(Q=\frac{[NO]^{2}}{[N_{2}][O_{2}]}\), when \(N_{2}\) and \(O_{2}\) are \(0\), \(Q
ightarrow\infty\)). Since \(K = 0.0093\lt1\), the forward reaction is not favored. But when starting with \(NO\), the reverse reaction \(2NO(g)
ightleftharpoons N_{2}(g)+O_{2}(g)\) will occur. The equilibrium constant for the reverse reaction \(K_{reverse}=\frac{1}{K}=\frac{1}{0.0093}\approx108\gt1\). So, at equilibrium, there will be very little \(NO\).

Step2: Calculate \(K\) for \(2NO(g)

ightleftharpoons N_{2}(g)+O_{2}(g)\)
For a reaction \(aA + bB
ightleftharpoons cC + dD\), \(K=\frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}\). If we reverse a reaction \(A
ightleftharpoons B\) (with \(K_1\)), the new equilibrium constant \(K_2=\frac{1}{K_1}\).
Given \(K_1 = 0.0093\) for \(N_{2}(g)+O_{2}(g)
ightleftharpoons 2NO(g)\), for \(2NO(g)
ightleftharpoons N_{2}(g)+O_{2}(g)\), \(K=\frac{1}{0.0093}\approx110\) (rounded to 2 significant digits \(1.1\times 10^{2}\))

Step3: Calculate \(K\) for \(3N_{2}(g)+3O_{2}(g)

ightleftharpoons 6NO(g)\)
If we multiply a reaction \(N_{2}(g)+O_{2}(g)
ightleftharpoons 2NO(g)\) (with \(K = 0.0093\)) by \(n = 3\), the new equilibrium constant \(K'=K^{n}\).
So \(K=(0.0093)^{3}=0.0093\times0.0093\times0.0093\approx8.0\times 10^{-7}\)

Answer:

• For the first question: There will be very little \(NO\).
• For \(2NO(g)

ightleftharpoons N_{2}(g)+O_{2}(g)\), \(K = 1.1\times 10^{2}\)

• For \(3N_{2}(g)+3O_{2}(g)

ightleftharpoons 6NO(g)\), \(K = 8.0\times 10^{-7}\)