If a graph is plotted between `log (a-x)` and `t`, the slope of the straight line is equal to `-0.03`. The specific reaction rate will be

Options

A. `6.9 xx 10^(-2)`
B. `6.9`
C. `0.69`
D. `6.9 xx 10^(-4)`

Explanation

To find the specific reaction rate (k) from the given information about the graph of log(a - x) versus time (t), we can follow these steps:

### Step 1: Understand the relationship
The relationship between the concentration of reactants and time for a first-order reaction can be expressed as:
\[ \log(a - x) = \log a - \frac{k}{2.303} t \]
Where:
- \( a \) is the initial concentration,
- \( x \) is the amount of reactant that has reacted,
- \( k \) is the specific reaction rate,
- \( t \) is time.

### Step 2: Identify the slope
From the equation, we can see that the slope (m) of the graph of log(a - x) versus time (t) is given by:
\[ m = -\frac{k}{2.303} \]
Given that the slope of the line is -0.03, we can set up the equation:
\[ -\frac{k}{2.303} = -0.03 \]

### Step 3: Solve for k
To find the specific reaction rate \( k \), we can rearrange the equation:
\[ k = 2.303 \times 0.03 \]

### Step 4: Calculate k
Now, we can calculate the value of \( k \):
\[ k = 2.303 \times 0.03 = 0.06909 \]

### Step 5: Express k in appropriate units
The unit of the specific reaction rate for a first-order reaction is typically expressed in \( \text{s}^{-1} \). Therefore, we can write:
\[ k \approx 0.0691 \, \text{s}^{-1} \]

### Final Answer
The specific reaction rate \( k \) is approximately \( 0.0691 \, \text{s}^{-1} \).

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