The correct plot for In k vs 1/T?

Options

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Answer

To solve the question regarding the correct plot for ln k versus 1/T, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Arrhenius Equation**:
The Arrhenius equation relates the rate constant (k) of a reaction to the temperature (T) and is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin.

2. **Take the Natural Logarithm**:
To plot ln k versus 1/T, we first take the natural logarithm of both sides of the Arrhenius equation:
\[ \ln k = \ln A - \frac{E_a}{RT} \]

3. **Rearranging the Equation**:
Rearranging the equation gives:
\[ \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \]
This equation is in the form of \( y = mx + c \) where:
- \( y = \ln k \)
- \( m = -\frac{E_a}{R} \) (the slope)
- \( x = \frac{1}{T} \)
- \( c = \ln A \) (the y-intercept)

4. **Identifying the Plot**:
From the rearranged equation, we can see that if we plot ln k on the y-axis and 1/T on the x-axis, we will get a straight line. The slope of this line will be negative, indicating that as the temperature increases (1/T decreases), the natural logarithm of the rate constant (ln k) increases.

5. **Conclusion**:
Therefore, the correct plot for ln k versus 1/T is a straight line with a negative slope.