A plot of log (a-x) against time 't' is a straight line. This indicates that the reaction is of :

Options

A. zero order
B. first order
C. second order
D. third order

Explanation

To solve the question, we need to analyze the relationship between the concentration of a reactant and time for different orders of reactions. The question states that a plot of log(a - x) against time 't' is a straight line, and we need to determine what this indicates about the order of the reaction.

### Step-by-Step Solution:

1. **Understanding the Reaction Order**:
- For a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant. The general rate equation is:
\[ -\frac{dA}{dt} = k[A] \]
where \( A \) is the concentration of the reactant, \( k \) is the rate constant, and \( t \) is time.

2. **Integrating the Rate Equation**:
- We can integrate the rate equation from the initial concentration \( A \) to \( A - x \) (where \( x \) is the amount of reactant that has reacted):
\[ \int_{A}^{A-x} \frac{dA}{A} = -k \int_{0}^{t} dt \]
- This results in:
\[ \ln(A - x) - \ln(A) = -kt \]

3. **Rearranging the Equation**:
- We can rearrange the equation using properties of logarithms:
\[ \ln(A - x) = -kt + \ln(A) \]
- This can be rewritten in terms of base 10 logarithm:
\[ \log(A - x) = \frac{-kt}{2.303} + \log(A) \]

4. **Identifying the Linear Relationship**:
- The equation now resembles the equation of a straight line \( y = mx + c \), where:
- \( y = \log(A - x) \)
- \( x = t \)
- \( m = \frac{-k}{2.303} \) (slope)
- \( c = \log(A) \) (y-intercept)

5. **Conclusion**:
- Since the plot of \( \log(A - x) \) against time \( t \) is a straight line, it indicates that the reaction follows first-order kinetics.

### Final Answer:
The reaction is of **first order**.

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