On introfucing a catalyst at `500K`, the rate of a first order reaction increases by `1.718` times. The activation energy in the presence of a catalyst is `4.15 KJ mol^(-1)`. The slope of the polt of `k(s^(-1))` against `1//T` in the absence of catalyst is

Options

A. `+1`
B. `-1`
C. `+1000`
D. `-1000`

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript and apply the relevant equations.

### Step 1: Understand the relationship between rate constants
The problem states that the rate of a first-order reaction increases by 1.718 times upon introducing a catalyst. This can be expressed in terms of the rate constants (k) for the reaction in the presence (k2) and absence (k1) of the catalyst:
\[ \frac{k_2}{k_1} = 1.718 \]

### Step 2: Use the Arrhenius equation
The Arrhenius equation relates the rate constant to the activation energy (Ea) and temperature (T):
\[ k = A e^{-\frac{E_a}{RT}} \]
Taking the logarithm of both sides gives:
\[ \ln k = \ln A - \frac{E_a}{RT} \]
For two different conditions (with and without catalyst), we can write:
\[ \ln k_2 - \ln k_1 = -\frac{E_{a, \text{catalyst}}}{RT} + \frac{E_{a, \text{no catalyst}}}{RT} \]
This simplifies to:
\[ \ln \frac{k_2}{k_1} = -\frac{(E_{a, \text{no catalyst}} - E_{a, \text{catalyst}})}{RT} \]

### Step 3: Substitute the known values
We know:
- \(E_{a, \text{catalyst}} = 4.15 \, \text{kJ/mol} = 4150 \, \text{J/mol}\)
- \(R = 8.314 \, \text{J/(mol K)}\)
- \(T = 500 \, \text{K}\)

Now we can substitute the values into the equation:
\[ \ln(1.718) = -\frac{(E_{a, \text{no catalyst}} - 4150)}{8.314 \times 500} \]

### Step 4: Calculate \(\ln(1.718)\)
Calculating \(\ln(1.718)\):
\[ \ln(1.718) \approx 0.235 \]

### Step 5: Rearranging the equation
Substituting \(\ln(1.718)\) into the equation gives:
\[ 0.235 = -\frac{(E_{a, \text{no catalyst}} - 4150)}{8.314 \times 500} \]
Multiplying both sides by \(-8.314 \times 500\):
\[ -0.235 \times 8.314 \times 500 = E_{a, \text{no catalyst}} - 4150 \]

### Step 6: Calculate the left-hand side
Calculating the left-hand side:
\[ -0.235 \times 8.314 \times 500 \approx -976.225 \]
Thus, we have:
\[ E_{a, \text{no catalyst}} - 4150 = -976.225 \]
So,
\[ E_{a, \text{no catalyst}} = 4150 - 976.225 \approx 3173.775 \, \text{J/mol} \approx 3.174 \, \text{kJ/mol} \]

### Step 7: Find the slope of the plot of k against 1/T
The slope (M) of the plot of \(\ln k\) against \(1/T\) is given by:
\[ M = -\frac{E_a}{R} \]
Substituting \(E_{a, \text{no catalyst}} = 3173.775 \, \text{J/mol}\):
\[ M = -\frac{3173.775}{8.314} \approx -381.1 \]

### Step 8: Final answer
The slope of the plot of \(k\) against \(1/T\) in the absence of catalyst is approximately \(-381.1\).