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.
Correctly label the speciation diagram below:
Match the following and choose the correct option from the following:
Note: The symbol indicates a stirrer for mixing (not to scale).
List I | List II | ||
---|---|---|---|
A | Mesozoic Era | I | Lower invertebrates |
B | Proterozoic Era | II | Fish & Amphibia |
C | Cenozoic Era | III | Birds & Reptiles |
D | Paleozoic Era | IV | Mammals |