Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 5 - Thermochemistry - Exercises: 5.56a


24.075 kJ/mol

Work Step by Step

There are 60g of water in the calorimeter and the temperature drops from 22 to 16.9C. First we find the $\Delta T$ in degrees C, which is the same as K. To do so we have: $\Delta T = 22 - 16.9 = 5.1 K $. To ind the heat released we use the specific heat of water, which is 4.18 J/g*K. Then we have: 4.18 J/g*K * 60g * 5.1K = 1279.08 J We then convert this to kJ as follows: 1.279 kJ To find the value of $\Delta H$ in kJ/mol we need to convert the 4.25g to moles of NH4NO3. We need to find the molar mass of NH4NO3 as follows: N is 14g/mol H is 1g/mol so 4*1 = 4 N is 14g/mol O is 16 g/mol, 3*16 = 48 14 + 4 + 14 + 48 = 80g/mol Therefore, 4.25g would be 4.25g * (1mol/80g) = 0.053 mol. Then we take the 1.279kJ and divide by the 0.053mol and we get 1.279kJ/ 0.053mol = 24.075 kJ/mol
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