How much heat (kJ)is required to convert 3.12 moles of liquid benzene at 75.1°C to gaseous benzene at 115.1°C? The following information may be useful. b.p = 80.1ᴼC C_m(liquid benzene)= 136.0 J/ (mol * °C) ΔH_vap = 30.72 kJ/mol C_m(gaseous benzene)= 36.6 J/ (mol * °C)

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To calculate the total heat required, we need to consider three steps: heating the liquid benzene from 75.1°C to its boiling point (80.1°C), vaporizing the liquid benzene at its boiling point, and then heating the gaseous benzene from 80.1°C to 115.1°C. 1. Heating the liquid benzene to its boiling point:$$ q_1 = m \cdot C_{\text{liquid}} \cdot \Delta T = 3.12 \, \text{mol} \cdot 136.0 \, \frac{\text{J}}{\text{mol} \cdot \text{°C}} \cdot (80.1°C - 75.1°C) $$$$ q_1 = 3.12 \cdot 136.0 \cdot 5.0 = 2121.6 \, \text{J} $$2. Vaporizing the liquid benzene at its boiling point:$$ q_2 = m \cdot \Delta H_{\text{vap}} = 3.12 \, \text{mol} \cdot 30.72 \, \frac{\text{kJ}}{\text{mol}} $$$$ q_2 = 3.12 \cdot 30.72 = 95.8464 \, \text{kJ} $$3. Heating the gaseous benzene to the final temperature:$$ q_3 = m \cdot C_{\text{gas}} \cdot \Delta T = 3.12 \, \text{mol} \cdot 36.6 \, \frac{\text{J}}{\text{mol} \cdot \text{°C}} \cdot (115.1°C - 80.1°C) $$$$ q_3 = 3.12 \cdot 36.6 \cdot 35.0 = 4009.44 \, \text{J} = 4.00944 \, \text{kJ} $$Total heat required:$$ q_{\text{total}} = q_1 + q_2 + q_3 $$$$ q_{\text{total}} = 2.1216 \, \text{kJ} + 95.8464 \, \text{kJ} + 4.00944 \, \text{kJ} $$$$ q_{\text{total}} \approx 101.97744 \, \text{kJ} $$Rounding to the nearest whole number, the total heat required is approximately 102.0 kJ.

How Much Heat (KJ)is Required to Convert 3