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 ( 4.79 \times 10^{-2} ) moles of ( mathrm{no}(mathrm{g}), 4.21 \times 10^{-2} ) moles of ( mathrm{br}_{2}(mathrm{~g}) ) and ( 8.93 \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 \(V = 1.00\space L\), \(c(NO)=4.79\times 10^{-2}\space M\), \(c(Br_{2})=4.21\times 10^{-2}\space M\), \(c(NOBr)=8.93\times 10^{-2}\space M\)

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

\(Q_{c}=\frac{(8.93\times 10^{-2})^{2}}{(4.79\times 10^{-2})^{2}\times(4.21\times 10^{-2})}\)
\(=\frac{7.97449\times 10^{-3}}{(2.29441\times 10^{-3})\times(4.21\times 10^{-2})}\)
\(=\frac{7.97449\times 10^{-3}}{9.6694661\times 10^{-5}}\)
\(Q_{c}\approx82.5\)

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

Given \(K_{c} = 154\) and \(Q_{c}=82.5\). Since \(Q_{c}

Answer:

The reaction quotient, \(Q_{c}\), equals \(82.5\). The reaction \(A. must run in the forward direction to reach equilibrium\)