Which of the following curves between log K and `1/T` is correct?

Options

A.
B.
C.
D.

Answer

To solve the question regarding the correct curve between log K and \( \frac{1}{T} \), we can follow these steps:

### Step 1: Understand the Arrhenius Equation
The Arrhenius equation describes how the rate constant \( k \) of a chemical reaction depends on temperature \( T \):
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) = rate constant
- \( A \) = pre-exponential factor (frequency factor)
- \( E_a \) = activation energy
- \( R \) = universal gas constant
- \( T \) = absolute temperature in Kelvin

### Step 2: Take the Logarithm
Taking the logarithm of both sides gives:
\[ \log k = \log A - \frac{E_a}{2.303RT} \]
This can be rewritten as:
\[ \log k = \log A - \frac{E_a}{R} \cdot \frac{1}{T} \]

### Step 3: Identify the Form of the Equation
The equation \( \log k = \log A - \frac{E_a}{R} \cdot \frac{1}{T} \) is in the form of a straight line \( y = mx + c \), where:
- \( y = \log k \)
- \( x = \frac{1}{T} \)
- \( m = -\frac{E_a}{R} \) (slope)
- \( c = \log A \) (y-intercept)

### Step 4: Plot the Graph
When we plot \( \log k \) on the y-axis against \( \frac{1}{T} \) on the x-axis, we will get a straight line with:
- A negative slope (since \( E_a \) is positive)
- A y-intercept at \( \log A \)

### Step 5: Determine the Correct Curve
Since the relationship is linear with a negative slope, the correct curve will be a straight line that slopes downwards from left to right.

### Conclusion
Thus, the correct option among the given choices is the one that represents a downward sloping straight line when plotting \( \log k \) against \( \frac{1}{T} \).

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