Arrhenius equation is:

Options

A. `d(Ink)/(dT) = (DeltaE^(+))/(RT)`
B. `(dInK)/(dT)=(DeltaE^(+))/(RT)^2`
C. `(dInK)/(dT)=-(DeltaE^(+))/(RT)^2`
D. `(dInK)/(dT)=-(DeltaE^(+))/(RT)`

Answer

To solve the question regarding the Arrhenius equation, we will follow these steps:

### Step 1: Write 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 temperature in Kelvin.

### Step 2: Take the Natural Logarithm
Taking the natural logarithm of both sides gives:
\[ \ln k = \ln A - \frac{E_a}{RT} \]

### Step 3: Differentiate with Respect to Temperature (T)
Now, we differentiate the equation with respect to \( T \):
- The term \( \ln A \) is a constant, so its derivative is 0.
- The term \( -\frac{E_a}{RT} \) needs to be differentiated.

Using the quotient rule:
\[ \frac{d}{dT}\left(-\frac{E_a}{RT}\right) = -E_a \cdot \frac{d}{dT}\left(\frac{1}{RT}\right) \]

The derivative of \( \frac{1}{RT} \) with respect to \( T \) is:
\[ \frac{d}{dT}\left(\frac{1}{T}\right) = -\frac{1}{T^2} \]
Thus,
\[ \frac{d}{dT}\left(-\frac{E_a}{RT}\right) = -E_a \cdot \left(-\frac{1}{R T^2}\right) = \frac{E_a}{RT^2} \]

### Step 4: Write the Final Expression
So, we have:
\[ \frac{d \ln k}{dT} = \frac{E_a}{RT^2} \]

### Conclusion
This expression indicates how the natural logarithm of the rate constant \( k \) changes with temperature \( T \).

### Final Answer
The Arrhenius equation leads to the result:
\[ \frac{d \ln k}{dT} = \frac{E_a}{RT^2} \]

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