For a certain reaction the variation of rate constant with temperature is given by the equation
`ln k_(t) = lnk_(0) + ((ln 3)t)/(10) (t ge 0^(@)C)`
The value of temperature coefficient of the reaction is

Options

A. 0.1
B. `1.0`
C. 10
D. 3

Explanation

To solve the problem, we need to find the temperature coefficient (μ) of the reaction using the given equation for the variation of the rate constant (k) with temperature (t).

### Step-by-Step Solution:

1. **Understand the Given Equation**:
The equation provided is:
\[ \ln k_t = \ln k_0 + \frac{\ln 3}{10} t \]
Here, \( k_t \) is the rate constant at temperature \( t \) and \( k_0 \) is the rate constant at 0°C.

2. **Recall the Relationship for Temperature Coefficient**:
The temperature coefficient (μ) is defined by the relationship:
\[ \frac{k_t}{k_0} = \mu^{(t/10)} \]
Taking the natural logarithm of both sides gives:
\[ \ln k_t - \ln k_0 = \frac{t}{10} \ln \mu \]

3. **Set Up the Comparison**:
From the given equation, we can rewrite it as:
\[ \ln k_t - \ln k_0 = \frac{\ln 3}{10} t \]
Now we have two expressions for \( \ln k_t - \ln k_0 \):
- From the given equation: \( \frac{\ln 3}{10} t \)
- From the temperature coefficient relationship: \( \frac{t}{10} \ln \mu \)

4. **Equate the Two Expressions**:
Since both expressions are equal, we can set them equal to each other:
\[ \frac{\ln 3}{10} t = \frac{t}{10} \ln \mu \]

5. **Simplify the Equation**:
We can cancel \( \frac{t}{10} \) from both sides (assuming \( t \neq 0 \)):
\[ \ln 3 = \ln \mu \]

6. **Solve for μ**:
Exponentiating both sides gives:
\[ \mu = 3 \]

### Conclusion:
The temperature coefficient (μ) of the reaction is 3.