Let's break down the problem step by step: **Given:** - Initial temperature, \(T_0 = 373\,K\) - Reaction: \(A \to B\) - Heat of reaction, \(\Delta H_{rxn} = -1\,kJ/mol = -1000\,J/mol\) (at 373 K) - Heat capacities: \(C_p(A) = C_p(B) = 50\,J/(mol\cdot K)\) (assume constant) - Conversion, \(X = 0.5\) (i.e., 50% of A converted to B) - Reactor is **adiabatic** (no heat exchange with surroundings) We are to find: **Temperature rise (\(\Delta T\)) after 50% conversion.** --- ## Step 1: Energy Balance for Adiabatic Batch Reactor For an adiabatic reactor (no heat loss): \[ \text{Total enthalpy change} = 0 \] \[ \Delta H_{rxn} \cdot n_{A,0} \cdot X + \int_{T_0}^{T} \left[ n_A C_p^A + n_B C_p^B \right] dT = 0 \] But since the heat capacities are the same and are constant, and since we're only interested in per mole basis, the equation simplifies. Let’s do per **mole of A initially present**. Let: - \(n_{A,0} = 1\) mol (assume 1 mol for calculation) - After conversion \(X\), \(n_A = n_{A,0}(1 - X)\), \(n_B = n_{A,0}X\) ## Step 2: Write the Enthalpy Balance The **change in enthalpy** (relative to the initial state) is: \[ \text{Change in enthalpy due to reaction:} = (\text{moles of A reacted}) \times (\Delta H_{rxn}) = n_{A,0} X \times \Delta H_{rxn} \] \[ \text{Change in enthalpy due to temperature rise:} \] Let the temperature after conversion be \(T\). The initial enthalpy (\(H_{initial}\)) = enthalpy of 1 mol A at \(T_0\). The final enthalpy (\(H_{final}\)) = enthalpy of \((1-X)\) mol A at \(T\) + \(X\) mol B at \(T\). So, the temperature rise \((T - T_0)\) is due to the heat released by the reaction. The total heat capacity of the system (sum for A and B): \[ C_{p,tot} = (1-X)C_p^A + X C_p^B = [(1-X) + X]C_p = C_p\ \text{(since both are equal)} \] So, total heat capacity per mole stays the same: \(C_p\). ## Step 3: Set Up the Equation The enthalpy balance: \[ (\text{heat released by reaction}) + (\text{heat absorbed by system as temperature rises}) = 0 \] \[ n_{A,0} X \Delta H_{rxn} + n_{A,0} C_p (T - T_0) = 0 \] Since \(n_{A,0} = 1\) mol, \[ X \Delta H_{rxn} + C_p (T - T_0) = 0 \] \[ C_p (T - T_0) = - X \Delta H_{rxn} \] \[ T - T_0 = \frac{- X \Delta H_{rxn}}{C_p} \] ## Step 4: Substitute the Values Given: - \(X = 0.5\) - \(\Delta H_{rxn} = -1000\,J/mol\) - \(C_p = 50\,J/(mol\cdot K)\) \[ T - T_0 = \frac{- (0.5) \times (-1000)}{50} \] \[ = \frac{0.5 \times 1000}{50} \] \[ = \frac{500}{50} \] \[ = 10\,K \] **Temperature rise is 10 K.** Since \(1\,K = 1\,^\circ C\), the temperature rise is **10°C**. --- ## **Final Answer** **(b) \(10^\circ C\)** --- **Summary Table:** | Step | Equation / Value | |----------------------|------------------------------------------------| | Energy balance | \(C_p (T-T_0) = -X\Delta H_{rxn}\) | | Substitute values | \(T - T_0 = \frac{-0.5 \times (-1000)}{50}\) | | Calculate | \(T - T_0 = 10\,K = 10^\circ C\) | | **Answer** | **(b) \(10^\circ C\)** |

Let's analyze why the other options are incorrect and confirm why 10°C is the correct temperature rise after 50% conversion. --- ### Why the other options are wrong: #### 1. **Option (a) 5°C** - The calculation for temperature rise is directly proportional to the heat released per mole of reactant and inversely proportional to the heat capacity. - Our calculation shows a 10°C rise, not 5°C. - The discrepancy likely arises from halving the heat release or miscalculating the heat capacity effect. Since the actual heat release (X × ΔH) is 500 J, and heat capacity is 50 J/K, the rise is 10°C, not 5°C. #### 2. **Option (c) 20°C** - The calculation indicates 10°C, so 20°C would be double the correct value. - To get 20°C, the heat released per mole would need to be 1000 J (double the actual 500 J), which isn't supported by the given data. #### 3. **Option (d) 100°C** - This is significantly higher than the calculated 10°C. - To reach a 100°C rise, the heat released would have to be 5,000 J, which is five times the actual heat release (X × ΔH = 500 J). - Such a large temperature rise is inconsistent with the given heat of reaction and heat capacity. --- ### **Summary of why 10°C is correct:** - The heat released during reaction at 50% conversion: \(0.5 \times 1000\,J = 500\,J\). - The heat capacity of the system: 50 J/K. - Temperature rise: \(\Delta T = \frac{\text{Heat released}}{\text{Heat capacity}} = \frac{500\,J}{50\,J/K} = 10\,K\). --- ### **Final note:** The key steps are: - Recognize the system's energy balance. - Use the given heat of reaction and conversion. - Apply the constant heat capacity. - Calculate the temperature rise as the ratio of heat released to heat capacity. **Therefore, the correct answer is:** # **(b) 10°C**