The heat of solution of ammonium nitrate is 26.2 kJ/mol. If a 5.368 g sample of NH₄NO₃ is added to 40.0 mL of water in a calorimeter at 23.5°C, what is the final temperature of the solution? The specific heat of water is 4.18 J/g·°C and the heat capacity of the calorimeter is 0.650 kJ/°C. A) 14.3°C B) 20.8°C C) -7.7°C D) 25.6°C E) 21.4°C

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Quizplus · Accepted Answer

First, calculate the moles of ammonium nitrate (NH4NO3) using its molar mass (80.04 g/mol). Moles = 5.368 g / 80.04 g/mol = 0.0671 mol. The total heat absorbed or released is q = moles × heat of solution = 0.0671 mol × 26.2 kJ/mol = 1.759 kJ. Since the heat of solution is positive, the process is endothermic, and heat is absorbed from the surroundings, causing the temperature to decrease. Convert q to J (1.759 kJ = 1759 J). The total mass of water is 40.0 mL × 1 g/mL = 40.0 g. Use q = mcΔT to find the temperature change, where m is the mass of water, c is the specific heat capacity, and ΔT is the temperature change. Rearrange to ΔT = q / (mc + calorimeter's heat capacity). The calorimeter's heat capacity must be converted to J/°C (0.650 kJ/°C = 650 J/°C). ΔT = 1759 J / ((40.0 g × 4.18 J/g°C) + 650 J/°C) = -2.1°C. The final temperature is the initial temperature minus the temperature change (since the solution absorbs heat): 23.5°C - 2.1°C = 21.4°C.

The Heat of Solution of Ammonium Nitrate Is 26