   # Adding 5.44 g of NH 4 NO 3 (s) to 150.0 g of water in a coffee-cup calorimeter (with stirring to dissolve the salt) resulted in a decrease in temperature from 18.6 °C to 16.2 °C. Calculate the enthalpy change for dissolving NH 4 NO 3 (s) in water, in kJ/mol. Assume the solution (whose mass is 155.4 g) has a specific heat capacity of 4.2 J/g · K. (Cold packs take advantage of the fact that dissolving ammonium nitrate in water is an endothermic process.) ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 5, Problem 41PS
Textbook Problem
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## Adding 5.44 g of NH4NO3(s) to 150.0 g of water in a coffee-cup calorimeter (with stirring to dissolve the salt) resulted in a decrease in temperature from 18.6 °C to 16.2 °C. Calculate the enthalpy change for dissolving NH4NO3(s) in water, in kJ/mol. Assume the solution (whose mass is 155.4 g) has a specific heat capacity of 4.2 J/g · K. (Cold packs take advantage of the fact that dissolving ammonium nitrate in water is an endothermic process.)

Interpretation Introduction

Interpretation:

The enthalpy change for the reaction for dissolving NH4NO3 in water has to be determined.

Concept Introduction:

Standard enthalpy of the reaction,ΔrHo, is the change in enthalpy that happens when matter is transformed by a given chemical reaction, when all reactants and products are in their standard states.

Enthalpy of the reaction,ΔrH is the change in enthalpy that happens when matter is transformed by a given chemical reaction

Heat energy required to raise the temperature of 1g of substance by 1K..Energy gained or lost can be calculated using the below equation.

q=C×m×ΔT

Where, q= energy gained or lost for a given mass of substance (m), C =specific heat capacity, ΔT= change in temperature.

### Explanation of Solution

Given,

Specific heat capacity of the solution=4.20JK/g

Mass of NH4NO3=5.44g

Mass of water =150g

Determine the amount of NH4NO3

5.44g(1mol÷80.043g) = 0.0679mol

Assume qr+qsol=0

qsol  can be calculated from q=C×m×ΔT, as

qsol= (5.44g of NH4NO3 + 150g of water) × 4

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