(a) Initial concentrations yield \(Q_c = 0.185\), indicating the system will proceed forward to reach equilibrium.
(b) Initial pressures result in \(Q_p = 0.468\), suggesting the system will move forward to achieve equilibrium.
(c) For \(Q_c = 2.15 > K_c\), the system will shift in the reverse direction.
(d) Initial pressures yield \(Q_p = 84.15 > K_p\), causing the system to move backward toward equilibrium.
(e) Since the denominator is zero, the system is not at equilibrium, and the reaction will proceed until product concentrations are nonzero.
(f) Initial concentrations lead to \(Q_c = 50 > K_c\), indicating the system will shift backward to reach equilibrium.
(a) For the reaction \(2 \text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3 \text{H}_2(g)\) with initial concentrations \([\text{NH}_3]_i = 0.540 \, \text{M}\), \([\text{N}_2]_i = 0.165 \, \text{M}\), \([\text{H}_2]_i = 0.110 \, \text{M}\), and \(K_c = 16\), the reaction quotient (\(Q_c\)) is calculated as follows:
\[ Q_c = \frac{[\text{N}_2][\text{H}_2]^3}{[\text{NH}_3]^2} \]
Substitute the given initial concentrations:
\[ Q_c = \frac{(0.165)(0.110)^3}{(0.540)^2} \approx 0.185 \]
Since \(Q_c\) is less than \(K_c\), the system will proceed in the forward direction to reach equilibrium.
(b) For the same reaction with initial pressures \(\text{Pi(NH}_3) = 1.95 \, \text{atm}\), \(\text{Pi(N}_2) = 11.45 \, \text{atm}\), \(\text{Pi(H}_2) = 11.45 \, \text{atm}\), and \(K_p = 4.4 \times 10^4\), the reaction quotient (\(Q_p\)) is calculated similarly:
\[ Q_p = \frac{\text{Pi(N}_2)\text{Pi(H}_2)^3}{(\text{Pi(NH}_3)^2} \approx 0.468 \]
Since \(Q_p\) is less than \(K_p\), the system will proceed in the forward direction to reach equilibrium.
(c) For the reaction \(2 \text{SO}_3(g) \rightleftharpoons 2 \text{SO}_2(g) + \text{O}_2(g)\) with initial concentrations \([\text{SO}_3]_i = 2.15 \, \text{M}\), \([\text{SO}_2]_i = 2.15 \, \text{M}\), \([\text{O}_2]_i = 2.15 \, \text{M}\), and \(K_c = 0.23\):
\[ Q_c = \frac{[\text{SO}_2]^2[\text{O}_2]}{[\text{SO}_3]^2} \]
Substitute the given initial concentrations:
\[ Q_c = \frac{(2.15)^2(2.15)}{(2.15)^2} = 2.15 \]
Since \(Q_c\) is greater than \(K_c\), the system will proceed in the reverse direction to reach equilibrium.
(d) For the same reaction with initial pressures \(\text{Pi(SO}_3) = 0.115 \, \text{atm}\), \(\text{Pi(SO}_2) = 1.05 \, \text{atm}\), \(\text{Pi(O}_2) = 1.20 \, \text{atm}\), and \(K_p = 7.8 \, \text{atm}\):
\[ Q_p = \frac{(\text{Pi(SO}_2)^2 \cdot \text{Pi(O}_2)}{\text{Pi(SO}_3)^2} \]
Substitute the given initial pressures:
\[ Q_p = \frac{(1.05)^2 \cdot 1.20}{(0.115)^2} \approx 84.15 \]
Since \(Q_p\) is greater than \(K_p\), the system will proceed in the reverse direction to reach equilibrium.
(e) For the reaction \(2 \text{NO(g)} + \text{Cl}_2(g) \rightleftharpoons 2 \text{NOCl(g)}\) with initial pressures \(\text{Pi(NO)} = 1.30 \, \text{atm}\), \(\text{Pi(Cl}_2) = 1.30 \, \text{atm}\), \(\text{Pi(NOCl)} = 0 \, \text{atm}\), and \(K_p = 4.0 \times 10^3\):
\[ Q_p = \frac{(\text{Pi(NO)})^2}{(\text{Pi(NOCl)})^2} \]
Substitute the given initial pressures:
\[ Q_p = \frac{(1.30)^2}{(0)^2} \]
Since the denominator is zero, the system is not at equilibrium, and the reaction will proceed until the concentrations of products are nonzero.
(f) For the reaction \( \text{N}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{NO(g)} \) with initial concentrations \([\text{N}_2]_i = 0.100 \, \text{M}\), \([\text{O}_2]_i = 0.200 \, \text{M}\), \([\text{NO]_i} = 1.00 \, \text{M}\), and \(K_c = 0.050\):
\[ Q_c = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]} \]
Substitute the given initial concentrations:
\[ Q_c = \frac{(1.00)^2}{(0.100)(0.200)} = 50 \]
Since \(Q_c\) is greater than \(K_c\), the system will proceed in the reverse direction to reach equilibrium.