Question

use the references to access important values if needed for this question.
consider the following system at equilibrium where \\( \delta h^{\circ}=268 \mathrm{~kj} \\), and \\( k_{c}=5.10 \times 10^{-6} \\), at \\( 548 \mathrm{~k} \\).
\\( \mathrm{nh}_{4} \mathrm{cl}(s) \
ightarrow \mathrm{nh}_{3}(g)+\mathrm{hcl}(g) \\)
when some moles of \\( \mathrm{nh}_{4} \mathrm{cl}(s) \\) are removed from the equilibrium system at constant temperature:
the value of \\( k_{c} \\)
o increases
decreases
o remains the same
the value of \\( q_{c} \\)
o is greater than \\( k_{c} \\)
o is equal to \\( k_{c} \\)
o is less than \\( k_{c} \\)
the reaction must
o run in the forward direction to reestablish equilibrium.
o run in the reverse direction to reestablish equilibrium.
o remain the same. it is already at equilibrium.
the number of moles of \\( \mathrm{nh}_{3} \\) will
o increase
decrease
o remain the same

Explanation:

Step1: Effect on \(K_c\)

The equilibrium constant \(K_c\) only depends on temperature. Since the temperature is constant, \(K_c\) remains the same.

Step2: Effect on \(Q_c\)

The reaction quotient \(Q_c = [NH_3][HCl]\) (solid \(NH_4Cl\) is not included in \(Q_c\) or \(K_c\) expressions). Removing \(NH_4Cl(s)\) (a solid) does not change the concentrations of \(NH_3(g)\) and \(HCl(g)\) immediately. So \(Q_c=K_c\)

Step3: Reaction direction

Since \(Q_c = K_c\), the reaction is still at equilibrium.

Step4: Moles of \(NH_3\)

As the reaction remains at equilibrium (no shift), the number of moles of \(NH_3\) remains the same.

Answer:

• The value of \(K_c\): remains the same
• The value of \(Q_c\): is equal to \(K_c\)
• The reaction must: remain the same. It is already at equilibrium.
• The number of moles of \(NH_3\) will: remain the same