Question

consider the following system at equilibrium where (delta h^{circ}=16.1 mathrm{~kj} / mathrm{mol}), and (k_{c}=6.50 \times 10^{-3}), at (298 mathrm{~k}).

(2 mathrm{nobr}(mathrm{g})
ightleftharpoons 2 mathrm{no}(mathrm{g})+mathrm{br}_{2}(mathrm{~g}))

when 0.19 moles of (mathrm{nobr}(mathrm{g})) are added to the equilibrium system at constant temperature:

the value of (k_{c})

the value of (q_{c} quad 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 concentration of no will

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 \(Q_{c}=\frac{[NO]^{2}[Br_{2}]}{[NOBr]^{2}}\). When \(NOBr\) is added, the denominator in the \(Q_{c}\) expression increases. So \(Q_{c}

Step3: Direction of reaction

According to Le - Chatelier's principle, when \(Q_{c}

Answer:

The value of \(K_{c}\) remains the same. The value of \(Q_{c}