Question

use the references to access important values if needed for this question.
consider the following reaction where ( k_{c}=0.0180 ) at ( 698 mathrm{~k} ).
2 mathrm{hi}(g)
ightleftharpoons mathrm{h}_{2}(g)+mathrm{i}_{2}(g)
a reaction mixture was found to contain ( 0.283 ) moles of ( mathrm{hi}(g), 0.0299 ) moles of ( mathrm{h}_{2}(g) ), and ( 0.0370 ) moles of ( mathrm{i}_{2}(g) ), in a ( 1.00 ) liter container.
calculate ( q_{c} ).
q_{c}=
is the reaction at equilibrium? if not, what direction must it run in order to reach equilibrium?
othe reaction must run in the forward direction to reach equilibrium.
othe reaction must run in the reverse direction to reach equilibrium.
othe reaction is at equilibrium.

Explanation:

Step1: Write the expression for \(Q_c\)

For the reaction \(2HI(g)
ightleftharpoons H_2(g)+I_2(g)\), the reaction quotient \(Q_c=\frac{[H_2][I_2]}{[HI]^2}\). Since the volume of the container \(V = 1.00\space L\), the molarity \(M=\frac{n}{V}\), and \(n\) (moles) is equal to \(M\) (molarity).

Step2: Substitute the values into the \(Q_c\) expression

We are given \([HI]=0.283\space M\), \([H_2]=0.0299\space M\), and \([I_2]=0.0370\space M\).

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Step3: Compare \(Q_c\) and \(K_c\)

Given \(K_c = 0.0180\) and \(Q_c=0.0139\). Since \(Q_c

Answer:

\(Q_c = 0.0139\), The reaction must run in the forward direction to reach equilibrium.