Step 1: Use the Reaction Quotient \( Q_c \)
The reaction quotient \( Q_c \) is given by:
$$ Q_c = \frac{[B][C]}{[A]^2} $$
Substitute the given concentrations:
$$ Q_c = \frac{(2 \times 10^{-3})(2 \times 10^{-3})}{(2 \times 10^{-3})^2} = \frac{4 \times 10^{-6}}{4 \times 10^{-6}} = 1 $$
Step 2: Compare \( Q_c \) with \( K_c \)
The equilibrium constant \( K_c \) is given as \( 4 \times 10^{-3} \).
Since \( Q_c = 1 \) and \( K_c = 4 \times 10^{-3} \), we see that:
$$ Q_c > K_c $$
Step 3: Direction of the Reaction
When \( Q_c > K_c \), the reaction proceeds in the backward direction to achieve equilibrium.
Therefore, the reaction will shift to the left, and there is a tendency for the reaction to go in the backward direction.
Step 4: Conclusion
The correct answer is: Option (3): The reaction has a tendency to go in the backward direction.
A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is :
A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is :