Skip to main content
Ch.6 - Thermochemistry

Chapter 6, Problem 76

Instant cold packs used to ice athletic injuries on the field contain ammonium nitrate and water separated by a thin plastic divider. When the divider is broken, the ammonium nitrate dissolves according to the endothermic reaction: NH4NO3(s)¡NH4 + (aq) + NO3- (aq) In order to measure the enthalpy change for this reaction, 1.25 g of NH4NO3 is dissolved in enough water to make 25.0 mL of solution. The initial temperature is 25.8 °C and the final temperature (after the solid dissolves) is 21.9 °C. Calculate the change in enthalpy for the reaction in kJ. (Use 1.0 g > mL as the density of the solution and 4.18 J>g # °C as the specific heat capacity.)

Verified Solution
Video duration:
0m:0s
This video solution was recommended by our tutors as helpful for the problem above.
3188
views
1
comments
Was this helpful?

Video transcript

Hey, everyone. And welcome back to another video. Instant cold packs are used to ice athletic injuries on the field by separating a thin plastic divider. It contains ammonium nitrate and water. When the divider is broken, the ammonium nitrate dissolves according to the endothermic reaction. Ammonium nitrate dissociates into the ammonium cations and nitrate anions in the reaction. 2.55 g of ammonium nitrate is dissolved in enough of water to make 59 mL of solution. The initial temperature is 30.8 °C and the final temperature is 24.9 °C. Calculate the change in enthalpy for the reaction in kilojoules use a 1.0 g per milliliter as the density of the solution and 4.18 joules per gram per Celsius as a specific heat capacity. Therefore, answer choices. A 57.4 kilojoules per mole, B 7.4 kilojoules per mole, C 45.7 kilojoules per mole and D 49.9 kilojoules per mole. So let's solve this problem. And our first step is to remember that the heat of the solution is equal to CM delta T, which is the final temperature minus the initial temperature. And we can actually reduce this to c the density of the solution multiplied by the volume of the solution multiplied by the change in temperature. The reason why we're doing this is because we are given the density and volume. So our sea specific heat is 4.18 jules per gram per Celsius. Our density is a 1.0 g per milliliter. Now, what else do we have? Well, essentially we need to use our volume which is 59 0.0 milliliters. And we want to multiply that by the change in temperature. So in this case, our final temperature is 24.9 °C and we're subtracting the initial 30.8 °C. Now, what do we get? Well, we get negative 1000 455 jules. Now, let's remember that according to the law of energy conservation heat of the reaction, which is also equal to the entropy change of the reaction is equal to the negative Q of the solution, negative heat of the solution, which gives us a positive value of the previous result. Now, what do we need to do? Well, essentially we want to calculate the change in entropy per mole. So we're going to take delta H of the reaction. If we want to calculate per mole of substance, we essentially have to take care of the reaction and divide by the number of moles of our substance, which is NH four nl three. So let's take 1455 Jews. And now on the bottom of this fraction, we need to calculate the number of moles of ammonium nitrate. We're given 2.55 g. And we need to divide that by the molar mass of ammonium nitrate, which is 80.05 g per mole. And now we can calculate the result. We get our answer in joules. If we divide by 1000, we get 45.7 kg joules per mole. That would be our final answer. And if we consider our answer choices, we can say that c the correct answer to this problem. Thank you for watching.
Related Practice
Textbook Question

When 0.514 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.8 °C to 29.4 °C. Find ΔErxn for the combustion of biphenyl in kJ>mol biphenyl. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.86 kJ/°C.

4680
views
4
rank
Textbook Question
Mothballs are composed primarily of the hydrocarbon naphthalene (C10H8). When 1.025 g of naphthalene burns in a bomb calorimeter, the temperature rises from 24.25 °C to 32.33 °C. Find ΔErxn for the combustion of naphthalene. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.11 kJ/°C.
3994
views
Textbook Question

Zinc metal reacts with hydrochloric acid according to the balanced equation: Zn(s) + 2 HCl(aq)¡ZnCl2(aq) + H2( g) When 0.103 g of Zn(s) is combined with enough HCl to make 50.0 mL of solution in a coffee-cup calorimeter, all of the zinc reacts, raising the temperature of the solution from 22.5 °C to 23.7 °C. Find ΔHrxn for this reaction as written. (Use 1.0 g>mL for the density of the solution and 4.18 J>g # °C as the specific heat capacity.)

27263
views
Textbook Question

For each generic reaction, determine the value of ΔH2 in terms of ΔH1. a. A + B → 2 C ΔH1 2 C→ A + B ΔH2 = ?

337
views
Textbook Question

For each generic reaction, determine the value of ΔH2 in terms of ΔH1. b. A + 1/2 B → C ΔH1 2 A + B → 2 C ΔH2 = ?

1908
views
Textbook Question

For each generic reaction, determine the value of ΔH2 in terms of ΔH1. c. A → B + 2 C ΔH1 1/2 B + C → 1/2 A ΔH2 = ?

372
views