Question

The hydrogenation of Ethyne is carried out at 600 K. The same reaction when carried out in the presence of a catalyst maintaining the same rate constant, the temperature required is only 400 K. If the catalyst lowers the Activation energy of the reaction by 20 kJ/mol, what is the value of \( E_a \)?

• A. 60 kJ/mol
• B. 100 kJ/mol
• C. 50 kJ/mol
• D. 80 kJ/mol

Hint:

When the rate constant is constant, the change in temperature affects the activation energy. Use the Arrhenius equation to calculate the change in activation energy for different temperatures.

Answer 1

The Correct Option is A

Step 1: Use the Arrhenius equation.
The Arrhenius equation relates the rate constant (\( k \)) to the temperature and activation energy (\( E_a \)) of a reaction:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor (constant),
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant (8.314 J/mol·K),
- \( T \) is the temperature in Kelvin.

Step 2: Relating the rate constants at two temperatures.
For two temperatures \( T_1 \) and \( T_2 \), the Arrhenius equation can be written as:
\[ \ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
where:
- \( k_1 \) and \( k_2 \) are the rate constants at temperatures \( T_1 \) and \( T_2 \), respectively.
Since the problem states that the rate constant remains the same before and after the catalyst is used, \( k_1 = k_2 \), which implies:
\[ \ln \left(\frac{k_2}{k_1}\right) = 0 \]
This simplifies the equation to: \[ 0 = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

Step 3: Apply the given values.
We are given:
- \( T_1 = 600 \, \text{K} \) (temperature without catalyst),
- \( T_2 = 400 \, \text{K} \) (temperature with catalyst),
- The catalyst lowers the activation energy by 20 kJ/mol.
Substitute these values into the equation:
\[ \ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{8.314} \left(\frac{1}{600} - \frac{1}{400}\right) \]
Simplify the equation: \[ 0 = \frac{E_a}{8.314} \left(\frac{1}{600} - \frac{1}{400}\right) \]

Step 4: Solve for \( E_a \).
Calculate the difference:
\[ \frac{1}{600} - \frac{1}{400} = \frac{2 - 3}{1200} = \frac{-1}{1200} \]
Thus, \[ 0 = \frac{E_a}{8.314} \left(\frac{-1}{1200}\right) \]
This gives: \[ E_a = 60 \, \text{kJ/mol} \]

Step 5: Conclusion.
Thus, the value of the activation energy \( E_a \) is 60 kJ/mol, which corresponds to option (A).