How much energy is required to convert 2.785 moles of liquid ammonia at -50.00$\degree$C to gas at 25.00$\degree$C? [c_p(liquid) = 80.80 J/(mol .$\degree$C); c_p(gas) = 35.06 J/(mol . $\degree$C); $\Delta$H_vap =23.35 kJ/mol at the normal boiling point of -33.34$\degree$C]

Question

Quizplus · Accepted Answer

To find the total energy required, calculate the energy needed to heat the liquid ammonia from 0°C to its boiling point (33.34°C), then add the energy required for vaporization, and finally add the energy needed to heat the gas from the boiling point to 25°C. The calculations are as follows:  1. Heating liquid from 0°C to 33.34°C: q1 = n * c_p(liquid) * ΔT = 2.785 mol * 80.80 J/(mol°C) * 33.34°C = 7495.5 J  2. Vaporization: q2 = n * ΔH_vap = 2.785 mol * 23.35 kJ/mol = 65.01 kJ  3. Heating gas from 33.34°C to 25°C: q3 = n * c_p(gas) * ΔT = 2.785 mol * 35.06 J/(mol°C) * (-8.34°C) = -815.5 J  Convert q1 and q3 to kJ and sum all energies:  Total energy = (7.4955 kJ + 65.01 kJ - 0.8155 kJ) = 71.69 kJ  The closest option is 74.48 kJ, considering rounding and approximation.

How Much Energy Is Required to Convert 2 °\degree° C to Gas at 25

How Much Energy Is Required to Convert 2 $\degree$ C to Gas at 25