Question

use the references to access important values if needed for this question.
consider the following reaction where ( k_{c}=7.00\times10^{-5} ) at 673 k.
( mathrm{nh}_{4}mathrm{i}(s)
ightleftharpoonsmathrm{nh}_{3}(g)+mathrm{hi}(g) )
a reaction mixture was found to contain 0.0570 moles of ( mathrm{nh}_{4}mathrm{i}(s) ), 0.00556 moles of ( mathrm{nh}_{3}(g) ), and 0.00837 moles of ( mathrm{hi}(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
must run in the forward direction to reach equilibrium.
must run in the reverse direction to reach equilibrium.
is at equilibrium.

Explanation:

Step1: Calculate the concentrations

Since the volume is \(1.00\) liter, the concentration of \(NH_3\) is \(c_{NH_3}=\frac{0.00556\space mol}{1.00\space L}= 0.00556\space M\) and the concentration of \(HI\) is \(c_{HI}=\frac{0.00837\space mol}{1.00\space L}=0.00837\space M\). The concentration of \(NH_4I\) (a solid) is not included in the reaction quotient expression.

Step2: Calculate the reaction quotient \(Q_c\)

The formula for \(Q_c\) for the reaction \(NH_4I(s)
ightleftharpoons NH_3(g)+HI(g)\) is \(Q_c = [NH_3][HI]\).
Substitute the values: \(Q_c=(0.00556)(0.00837)\)

$$Q_c = 4.65\times10^{-5}$$

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

Given \(K_c = 7.00\times 10^{-5}\). Since \(Q_c=4.65\times 10^{-5}

Answer:

The reaction quotient \(Q_c\) equals \(4.65\times 10^{-5}\). The reaction must run in the forward direction to reach equilibrium.