A sample of ammonium nitrate having a mass of 3.88 grams of water is dissolved in 60.0 g of water. The temperature of the water decreases from 23.0°C to 18.4°C. What is the molar enthalpy of dissolution for ammonium nitrate?

Question

Naomi Aronson · Accepted Answer

#+"23.8 kJ/mol"#

Even before doing any calculations, you can look at the change in temperature measured for the water to say that you can expect the standard molar enthalpy of dissolution of ammonium nitrate to be positive.

Since adding the ammonium nitrate salt to the water results in a decrease in temperature, you can conclude that the dissolution of ammonium nitrate absorbs heat from the surroundings #-># you're dealing with an endothermic reaction.

So, your strategy here will be to use the mass of the water, its specific heat, and its change in temperature to determine how much heat was absorbed by the dissolution reaction.

Your go-to equation will be

#color(blue)(q = m * c * DeltaT)" "#, where

#q# - heat absorbed/lost #m# - the mass of the sample #c# - the specific heat of water, equal to #4.18 "J"/("g" ""^@"C")# #DeltaT# - the change in temperature, defined as final temperature minus initial temperature

Plug in your values to get

#q = 60.0 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * (18.4 - 23.0)color(red)(cancel(color(black)(""^@"C")))#

#q = -"1153.7 J"#

Now, don't be confused by the negative sign. If you look at things from water's perspective, you can say that water gives off heat, hence the negative sign.

Now, the standard molar enthalpy of dissolution is usually expressed in kilojoules per mole. Use ammonium nitrate's molar mass to determine how many moles you get in that sample

#3.88 color(red)(cancel(color(black)("g"))) * ("1 mole NH"_4"NO"_3)/(80.04color(red)(cancel(color(black)("g")))) = "0.04848 moles NH"_4"NO"_3#

Now, if the dissolution of #0.04848# moles of ammonium nitrate required #"1153.7 J"# of heat, it follows that the dissolution of one mole of the compound will require

#1color(red)(cancel(color(black)("mole"))) * "1153.7 J"/(0.04848color(red)(cancel(color(black)("moles")))) = "23797.4 J"#

Expressed in kilojoules per mole and rounded to three sig figs, the answer will be

#DeltaH_"sol"^@ = color(green)(+"23.8 kJ/mol")#

Peyton Adams · Answer

undefined The molar enthalpy of dissolution for ammonium nitrate can be calculated using the formula:

$$ \Delta H_{\text{dissolution}} = \frac{q}{n} $$

Where:
- $ q $ is the heat absorbed or released by the solution (in joules)
- $ n $ is the number of moles of the solute (ammonium nitrate in this case)

The heat absorbed or released by the solution ($ q $) can be calculated using the formula:

$$ q = mc\Delta T $$

Where:
- $ m $ is the mass of the solvent (water) in grams
- $ c $ is the specific heat capacity of water (4.18 J/g°C)
- $ \Delta T $ is the change in temperature (final temperature - initial temperature) in °C

Once $ q $ is found, $ n $ can be calculated by dividing the mass of the solute (ammonium nitrate) by its molar mass. Then, $ \Delta H_{\text{dissolution}} $ can be determined.

Given:
- Mass of water (solvent), $ m = 60.0 $ g
- Change in temperature, $ \Delta T = 23.0°C - 18.4°C = 4.6°C $
- Specific heat capacity of water, $ c = 4.18 $ J/g°C
- Mass of ammonium nitrate (solute), $ m_{\text{NH}_4\text{NO}_3} = 3.88 $ g
- Initial and final temperatures

1. Calculate $ q $ using the heat capacity formula.
2. Calculate the number of moles of ammonium nitrate ($ n $) using its mass and molar mass.
3. Use the formula $ \Delta H_{\text{dissolution}} = \frac{q}{n} $ to find the molar enthalpy of dissolution.