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 \(\boxed{10°C}\).

Answer 2

This problem asks for the adiabatic temperature rise in a batch reactor once 50% of A has reacted, given the reaction is exothermic by 1 kJ/mol. We can check each option by working out how much sensible heat the reacting mixture must absorb from the heat released by the reaction, using a direct enthalpy balance instead of the shortcut formula.

Basis: 1 mol of A charged initially. Since the reactor is adiabatic, all the heat released by the reaction stays inside the reactor and raises the temperature of the reacting mixture (both A and B share the same heat capacity, so the total heat capacity of the mixture stays at \( 1 \times 50 = 50 \, \text{J/K} \) regardless of how far the reaction has proceeded). At a conversion of \( X = 0.5 \), the moles of A reacted equal 0.5 mol, so the heat liberated is: \[ Q = (-\Delta H_{rxn}) \times n_{reacted} = 1000 \, \text{J/mol} \times 0.5 \, \text{mol} = 500 \, \text{J} \] This heat raises the mixture temperature by \( \Delta T = Q / (n_{total} C_p) = 500 / 50 = 10 \, \text{K} \), i.e. a \( 10^{\circ}\text{C} \) rise. With this energy balance fixed, we can now check each option against it:

1. 5°C: This would require only 250 J of heat to be released, i.e. a conversion of just 0.25 at this heat of reaction, half of what is actually specified, so this undershoots the real energy release.
2. 10°C: This matches exactly the 500 J of heat computed above being absorbed by the 50 J/K total heat capacity of the mixture, so it is consistent with the given conversion and heat of reaction.
3. 20°C: This would require 1000 J of heat, corresponding to a full conversion of 1.0 (all of A reacted), not the stated 0.5, so it overstates the heat released at this conversion.
4. 100°C: This would require 5000 J of heat, five times what the reaction can supply at this heat of reaction and heat capacity even at complete conversion, clearly inconsistent with the numbers given.

Only the 10°C rise is consistent with the enthalpy balance for this adiabatic reactor at 50% conversion.

Therefore, the correct answer is 10°C.