Question

A batch adiabatic reactor at an initial temperature of 373K is being used for the reaction A to B. Assume the heat of reaction to be -1 kJ/mol at 373K and the heat capacity of both A and B to be constant and equal to 50J/mol-K. The temperature rise after a conversion of 0.5 will be

• A. 5°C
• B. 10°C
• C. 20°C
• D. 100°C

Hint:

In adiabatic reactions, the temperature change depends on the heat of reaction and the heat capacity of the system.

Answer 1

The Correct Option is B

Step 1: Understanding the problem.
In a batch adiabatic reactor, no heat is exchanged with the surroundings. Therefore, the temperature change is due to the heat released or absorbed during the reaction. The formula to calculate the temperature rise (\(\Delta T\)) is given by: \[ \Delta T = \frac{-\Delta H \cdot X}{C_p} \] Where: - \( \Delta H \) is the heat of reaction (\(-1 \, \text{kJ/mol}\)) - \( X \) is the conversion (0.5) - \( C_p \) is the heat capacity (50 J/mol-K) Step 2: Plugging in the values.
First, convert the heat of reaction to J/mol: \[ \Delta H = -1 \, \text{kJ/mol} = -1000 \, \text{J/mol} \] Now, calculate the temperature rise: \[ \Delta T = \frac{-(-1000 \, \text{J/mol}) \times 0.5}{50 \, \text{J/mol-K}} = \frac{500}{50} = 10°C \] Step 3: Conclusion.
The temperature rise after a conversion of 0.5 is 10°C. The correct answer is \(\boxed{10°C}\).