Rate constant k varies with temperature T(k) as given by equation
`logk=logA-(E_(a))/(2.303RT)`
A plot between log k and `(1)/(T)` is straight line from the graph we can say

Options

A. `E_(a)=2.303Rxx"slope"`
B. `E_(a)=2.303R^(2)xxtantheta`
C. `E_(a)=Rtantheta`
D. None of these

Explanation

To solve the question regarding the relationship between the rate constant \( k \) and temperature \( T \) as given by the equation:

\[ \log k = \log A - \frac{E_a}{2.303RT} \]

we can follow these steps:

### Step 1: Identify the equation form
The equation resembles the linear form \( y = mx + c \), where:
- \( y \) is \( \log k \)
- \( m \) is the slope of the line, which is \(-\frac{E_a}{2.303R}\)
- \( x \) is \(\frac{1}{T}\)
- \( c \) is \(\log A\)

### Step 2: Understand the slope
From the equation, we can see that the slope of the line plotted between \( \log k \) and \(\frac{1}{T}\) is equal to \(-\frac{E_a}{2.303R}\). This indicates that the slope is negative, which implies that as the temperature increases (and thus \(\frac{1}{T}\) decreases), the value of \(\log k\) increases.

### Step 3: Relate slope to activation energy
Since the slope is related to the activation energy \( E_a \), we can express the activation energy as:
\[ E_a = -\text{slope} \times 2.303R \]

### Step 4: Conclusion
From the graph, we can conclude that the activation energy \( E_a \) can be determined from the slope of the line. The negative slope indicates that the activation energy is a positive quantity, which is consistent with the physical interpretation of activation energy in chemical kinetics.

### Final Answer
Thus, from the graph, we can say that the activation energy \( E_a \) can be calculated using the slope of the line obtained from the plot of \( \log k \) versus \(\frac{1}{T}\).

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