Question

consider the following reaction where ( k_{c}=154 ) at ( 298 mathrm{~k} ).
( 2 mathrm{no}(mathrm{g})+mathrm{br}_{2}(mathrm{~g})
ightleftharpoons 2 mathrm{nobr}(mathrm{g}) )
a reaction mixture was found to contain ( 1.98 \times 10^{-2} ) moles of ( mathrm{no}(mathrm{g}), 4.29 \times 10^{-2} ) moles of ( mathrm{br}_{2}(mathrm{~g}) ) and ( 7.94 \times 10^{-2} ) moles of ( mathrm{nobr}(mathrm{g}) ), in a 1.00 liter container.
is the reaction at equilibrium?
if not, what direction must it run in order to reach equilibrium?
the reaction quotient, ( q_{c} ) equals
the reaction
a. must run in the forward direction to reach equilibrium.
b. must run in the reverse direction to reach equilibrium.
c. is at equilibrium.

Explanation:

Step1: Calculate concentrations

Since concentration \(c=\frac{n}{V}\) and \(V = 1.00L\), \([NO]=1.98\times 10^{-2}M\), \([Br_{2}]=4.29\times 10^{-2}M\), \([NOBr]=7.94\times 10^{-2}M\)

Step2: Write the formula for \(Q_{c}\)

For the reaction \(2NO(g)+Br_{2}(g)
ightleftharpoons 2NOBr(g)\), \(Q_{c}=\frac{[NOBr]^{2}}{[NO]^{2}[Br_{2}]}\)

Step3: Substitute values into \(Q_{c}\) formula

$$
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$$

Step4: Compare \(Q_{c}\) and \(K_{c}\)

Given \(K_{c}=154\) at \(298K\). Since \(Q_{c}(375)>K_{c}(154)\)

Answer:

The reaction quotient \(Q_{c}\) equals \(375\). The reaction B. must run in the reverse direction to reach equilibrium