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 solve the problem, we need to determine the specific reaction rate (rate constant, k) given the slope of the graph plotted between log(a - x) and time (t).

### Step-by-Step Solution:

1. **Understanding the Reaction Order**:
- The question states that a graph is plotted between log(a - x) and time (t).
- A straight line indicates that the reaction is first order.

2. **Using the First Order Reaction Equation**:
- For a first-order reaction, the relationship can be expressed as:
\[ \log(a - x) = \log a - \frac{k}{2.303} t \]
- Rearranging this gives us:
\[ \log(a - x) = -\frac{k}{2.303} t + \log a \]
- This is in the form of \(y = mx + c\), where \(y = \log(a - x)\), \(m = -\frac{k}{2.303}\), and \(c = \log a\).

3. **Identifying the Slope**:
- From the problem, we know the slope of the line (m) is given as -0.03.
- Therefore, we can set up the equation:
\[ -\frac{k}{2.303} = -0.03 \]

4. **Solving for the Rate Constant (k)**:
- To find k, we can rearrange the equation:
\[ \frac{k}{2.303} = 0.03 \]
- Multiplying both sides by 2.303 gives:
\[ k = 2.303 \times 0.03 \]

5. **Calculating k**:
- Performing the multiplication:
\[ k = 0.06909 \]
- This can be expressed in scientific notation:
\[ k \approx 6.9 \times 10^{-2} \]

6. **Conclusion**:
- The specific reaction rate (k) is approximately \(6.9 \times 10^{-2} \, \text{s}^{-1}\).

### Final Answer:
The specific reaction rate (k) is \(6.9 \times 10^{-2} \, \text{s}^{-1}\).