If a second order reaction is 75% complete in 1 hr , calculate the rate constant .

Answer

To solve the problem, we need to calculate the rate constant \( k \) for a second-order reaction that is 75% complete in 1 hour. Here's the step-by-step solution:

### Step 1: Understand the Reaction
In a second-order reaction, the rate of reaction depends on the concentration of the reactant squared. The general form of the integrated rate law for a second-order reaction is:

\[ \frac{1}{A} - \frac{1}{A_0} = kt \]

where:
- \( A_0 \) is the initial concentration,
- \( A \) is the concentration at time \( t \),
- \( k \) is the rate constant,
- \( t \) is the time.

### Step 2: Define Initial and Final Concentrations
Given that the reaction is 75% complete in 1 hour, we can define the concentrations as follows:
- Initial concentration \( A_0 = 100\% \) (or we can take it as 1 for simplicity).
- After 75% completion, the remaining concentration \( A = 100\% - 75\% = 25\% \) (or 0.25).

### Step 3: Substitute Values into the Rate Law
Now we can substitute the known values into the integrated rate law:

\[ \frac{1}{A} - \frac{1}{A_0} = kt \]

Substituting \( A_0 = 1 \) and \( A = 0.25 \):

\[ \frac{1}{0.25} - \frac{1}{1} = k \cdot 1 \text{ hour} \]

### Step 4: Calculate the Left Side
Calculate the left side of the equation:

\[ \frac{1}{0.25} = 4 \quad \text{and} \quad \frac{1}{1} = 1 \]

Thus:

\[ 4 - 1 = k \cdot 1 \]

This simplifies to:

\[ 3 = k \cdot 1 \]

### Step 5: Solve for \( k \)
Now we can solve for \( k \):

\[ k = 3 \text{ hr}^{-1} \]

### Step 6: Convert to Appropriate Units
Since the question asks for the rate constant, we typically express it in terms of \( \text{M}^{-1}\text{hr}^{-1} \). However, since we initially took concentrations as percentages, we can express it as:

\[ k = 3 \text{ M}^{-1}\text{hr}^{-1} \]

### Final Answer
The rate constant \( k \) for the second-order reaction that is 75% complete in 1 hour is:

\[ k = 3 \text{ M}^{-1}\text{hr}^{-1} \]

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