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

Question

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First, we need to calculate the energy required to raise the temperature of liquid ammonia from F1F1F1S1F1F1F1050.00°C to its normal boiling point of F1F1F1S1F1F1F1033.34°C: (80.80 J/(molF1F1F1S1F1F1F10°C)) x (2.785 mol) x (33.34°C) = 7500 J Next, we need to calculate the energy required to vaporize the liquid ammonia at its normal boiling point: (23.35 kJ/mol) x (2.785 mol) = 65.0185 kJ Finally, we need to calculate the energy required to cool the gaseous ammonia from its boiling point at F1F1F1S1F1F1F1033.34°C to 25.00°C: (35.06 J/(molF1F1F1S1F1F1F10°C)) x (2.785 mol) x (1008.34°C) = 9824.8 J Adding up all the required energies gives: 7500 J + 65.0185 kJ + 9824.8 J = 74.48 kJ. Therefore, the answer is A.

How Much Energy Is Required to Convert 2 Δ\DeltaΔ Hvap = 23

How Much Energy Is Required to Convert 2 $\Delta$ H_vap = 23