Question

Experiment 13-Post-Lab Assignment 1. Athletes use cold packs containing ammonium nitrate solutions to ice their injuries. A 1.25 G sample of ammonium nitrate is dissolved in 25.0 mL of water to form a solution. The temperature of the solution falls from 25.8°C to 21.9°C. Find the enthalpy of the reaction (AH ), in kJ/mol. Assume the density of water is 1.00 g/mL and the heat capacity of the total solution to be 4.184 J/g °C 2. Calculate AHⓇ in kJ/mol for the reaction: CH (9) + NH3(g) → HCN(9) + 3 H (9) N2(g) + 3 H2(9) ► NH3(9) AH = -91.8 kJ/mol C(s) + 2 H2(g) - CH (9) AH2 = -74.9 kJ/mol Ha(9) + 2 C(s) + N (9) + 2 HCN(g) AH, = + 270.3 kJ/mol 13-7

Answer 1

Here is the Solution:

1.25 g sample of ammonium nitrate dissolves in 25. mL of water to form a solution. mass of the solution = 25+ 1,25 = 26.25 g (m) • Temperature change= 25.8-21,9 = 3.9°C speutic heat = 4.184 J/g °C (6) (OT) Q = micioT Q = (26.25g7x (4.1843/80) x (3.90) Q = 428.337 J .Given mass of ammonium nitrate= 1.25 g Molar mass of ammonium nitrate = 80.043 g/mol given mass 1.25 g No. of moles= - 80.043 g/mol No. of moles = 0.015 6 166 mol molar mass The enthalpy of the reaction can be calculated as o Hron = No. of moles 428.337 ] a Hoon = 0.0156166 mol Ohran = 27428,31 J/mol 1 kJ/mol = 27428.31 IJmol X - 1000 Lmot 27.428 kJ/mol

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