K represents the rate constant of a reaction when log K is p lotted again st 1/T (T=temperature) the graph obtained is a

Options

A. curve
B. a straight line with a constant positive slope
C. a straight line with constant negative slope
D. a straight line with no slope

Explanation

To solve the problem, we need to analyze the relationship between the rate constant (K) and temperature (T) using the Arrhenius equation. Here’s a step-by-step solution:

### Step 1: Understand the Arrhenius Equation
The Arrhenius equation 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 absolute temperature.

### Step 2: Take the Logarithm of Both Sides
To analyze the relationship in terms of logarithms, we take the logarithm of both sides:
\[ \log K = \log A - \frac{E_a}{2.303RT} \]
This can be rearranged to:
\[ \log K = -\frac{E_a}{2.303R} \cdot \frac{1}{T} + \log A \]

### Step 3: Identify the Linear Form
The equation now resembles the linear form \( y = mx + c \):
- Let \( y = \log K \)
- Let \( x = \frac{1}{T} \)
- The slope \( m = -\frac{E_a}{2.303R} \)
- The y-intercept \( c = \log A \)

### Step 4: Analyze the Graph
From the equation, we can conclude:
- The graph of \( \log K \) versus \( \frac{1}{T} \) is a straight line.
- The slope of the line is negative because \( -\frac{E_a}{2.303R} \) is negative (since \( E_a \) and \( R \) are positive).
- The intercept is \( \log A \), which means the line does not pass through the origin.

### Conclusion
The graph obtained by plotting \( \log K \) against \( \frac{1}{T} \) is a straight line with a constant negative slope.

### Final Answer
The correct nature of the graph is: **A straight line with a constant negative slope.**

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