Rate of reaction in absence of catalyst at 700 K is same as in presence of catalyst at 500 K. If catalyst decreases activation energy barrier by 30 kJ/mole, determine activation energy in presence of catalyst. (Assume 'A' factor to be same in both cases)

Options

A. 75 kJ/mole
B. 135 KJ/mole
C. 105 KJ/mole
D. 125 KJ/mole

Explanation

To determine the activation energy in the presence of a catalyst (E_A1), we can use the information provided in the question along with the Arrhenius equation. Here’s a step-by-step solution:

### Step 1: Understand the Given Information
- The rate of reaction in the absence of a catalyst at 700 K is the same as in the presence of a catalyst at 500 K.
- The catalyst decreases the activation energy barrier by 30 kJ/mol.

### Step 2: Define Variables
Let:
- E_A2 = Activation energy in the absence of catalyst
- E_A1 = Activation energy in the presence of catalyst
- Given that E_A1 = E_A2 - 30 kJ/mol (since the catalyst decreases the activation energy by 30 kJ/mol)

### Step 3: Use the Arrhenius Equation
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_A}{RT}} \]
Where:
- k = rate constant
- A = pre-exponential factor (assumed to be the same for both cases)
- E_A = activation energy
- R = universal gas constant (8.314 J/mol·K)
- T = temperature in Kelvin

Since the rates (and thus the rate constants) are equal, we can set up the following equation:
\[ A e^{-\frac{E_A1}{RT_1}} = A e^{-\frac{E_A2}{RT_2}} \]
Where:
- T_1 = 500 K (temperature in presence of catalyst)
- T_2 = 700 K (temperature in absence of catalyst)

### Step 4: Simplify the Equation
Since A cancels out, we have:
\[ e^{-\frac{E_A1}{RT_1}} = e^{-\frac{E_A2}{RT_2}} \]

Taking the natural logarithm of both sides gives:
\[ -\frac{E_A1}{RT_1} = -\frac{E_A2}{RT_2} \]

Thus:
\[ \frac{E_A1}{T_1} = \frac{E_A2}{T_2} \]

### Step 5: Substitute Known Values
We know:
- T_1 = 500 K
- T_2 = 700 K
- E_A1 = E_A2 - 30

Substituting E_A1 in the equation:
\[ \frac{E_A2 - 30}{500} = \frac{E_A2}{700} \]

### Step 6: Cross Multiply and Solve for E_A2
Cross multiplying gives:
\[ 700(E_A2 - 30) = 500E_A2 \]
Expanding this:
\[ 700E_A2 - 21000 = 500E_A2 \]

Rearranging gives:
\[ 700E_A2 - 500E_A2 = 21000 \]
\[ 200E_A2 = 21000 \]
\[ E_A2 = \frac{21000}{200} = 105 \text{ kJ/mol} \]

### Step 7: Find E_A1
Now, substitute E_A2 back to find E_A1:
\[ E_A1 = E_A2 - 30 = 105 - 30 = 75 \text{ kJ/mol} \]

### Final Answer
The activation energy in the presence of the catalyst (E_A1) is **75 kJ/mol**.

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