Question

use the references to access important values if needed for this question. consider the following system at equilibrium where \\( \delta h^{\circ}=92.7 \mathrm{~kj} \\), and \\( k_{c}=1.80 \times 10^{-4} \\), at \\( 290 \mathrm{~k} \\). \\( \mathrm{nh}_{4} \mathrm{hs}(s) \
ightarrow \mathrm{nh}_{3}(g)+\mathrm{h}_{2} \mathrm{~s}(g) \\) when some moles of \\( \mathrm{nh}_{4} \mathrm{hs}(s) \\) are removed from the equilibrium system at constant temperature: the value of \\( k_{c} \\) oincreases odecreases oremains the same the value of \\( q_{c} \\) ois greater than \\( k_{c} \\) ois equal to \\( k_{c} \\) ois less than \\( k_{c} \\) the reaction must orun in the forward direction to reestablish equilibrium. orun in the reverse direction to reestablish equilibrium. oremain the same. it is already at equilibrium. the number of moles of \\( \mathrm{nh}_{3} \\) will oincrease odecrease oremain the same

Explanation:

• For \(K_{c}\): The equilibrium constant \(K_{c}\) only depends on temperature. Since the temperature is constant, \(K_{c}\) remains the same.
• For \(Q_{c}\): The expression for \(Q_{c}\) for the reaction \(NH_{4}HS(s)\to NH_{3}(g)+H_{2}S(g)\) is \(Q_{c}=[NH_{3}][H_{2}S]\). Removing \(NH_{4}HS(s)\) (a solid) does not change the concentrations of \(NH_{3}(g)\) and \(H_{2}S(g)\) immediately. So \(Q_{c}\) is equal to \(K_{c}\) initially. But since the amount of reactant (solid) is decreased, the system will try to maintain the equilibrium. However, because \(NH_{4}HS(s)\) is a solid and its concentration (activity) is considered as 1 (constant) in the equilibrium expression, removing some of it does not affect \(Q_{c}\) in terms of concentration - based calculation (\(Q_{c}\) formula only involves gases). So \(Q_{c}\) remains equal to \(K_{c}\).
• For the reaction direction: Since \(Q_{c}=K_{c}\), the reaction remains the same. It is already at equilibrium.
• For the number of moles of \(NH_{3}\): As the reaction remains at equilibrium (because \(Q_{c} = K_{c}\)), 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