In a reversible reaction, the forward reaction was `3` times faster than that of reverse reaction. The reaction quotient is.....

Answer

To solve the problem step by step, we can follow this approach:

### Step 1: Understand the Reaction
We have a reversible reaction where the forward reaction is faster than the reverse reaction. Let's denote the forward reaction as \( A \rightleftharpoons B \).

### Step 2: Define the Rate Constants
Let \( k_f \) be the rate constant for the forward reaction and \( k_b \) be the rate constant for the reverse reaction. According to the problem, the forward reaction is 3 times faster than the reverse reaction.

### Step 3: Set Up the Relationship
From the information given, we can express this relationship mathematically:
\[ k_f = 3k_b \]

### Step 4: Define the Reaction Quotient
The reaction quotient \( Q_c \) for a reaction at equilibrium is defined as the ratio of the concentrations of the products to the reactants, each raised to the power of their coefficients in the balanced equation. In terms of rate constants, we can express it as:
\[ Q_c = \frac{k_f}{k_b} \]

### Step 5: Substitute the Values
Now, substituting the value of \( k_f \) from Step 3 into the equation for \( Q_c \):
\[ Q_c = \frac{3k_b}{k_b} \]

### Step 6: Simplify the Expression
When we simplify the expression, the \( k_b \) terms cancel out:
\[ Q_c = 3 \]

### Conclusion
Thus, the reaction quotient \( Q_c \) for this reversible reaction is \( 3 \).

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