Question

use the references to access important values if needed for this question.
consider the following system at equilibrium where δh = 16.1 kj, and kc = 6.50 × 10^-3, at 298 k:
2nobr(g) ⇌ 2no(g) + br2(g)
if the volume on the equilibrium system is suddenly increased at constant temperature:
the value of kc
o increases
o decreases
o remains the same
the value of qc
o is greater than kc
o is equal to kc
o is less than kc
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 br2 will
o increase
o decrease
o remain 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 \(Q_{c}\): When the volume is increased, the concentrations of all gases decrease. The reaction has more moles of gas on the product side (\(2 + 1=3\) moles of products vs \(2\) moles of reactants). Using the formula \(Q_{c}=\frac{[NO]^{2}[Br_{2}]}{[NOBr]^{2}}\), a decrease in concentration (due to volume increase) will make \(Q_{c}
• For the reaction direction: Since \(Q_{c}
• For the number of moles of \(Br_{2}\): As the reaction shifts forward, more \(Br_{2}\) is formed, so the number of moles of \(Br_{2}\) increases.

Answer:

• The value of \(K_{c}\): remains the same
• The value of \(Q_{c}\): is less than \(K_{c}\)
• The reaction must: run in the forward direction to reestablish equilibrium
• The number of moles of \(Br_{2}\) will: increase