Question

A 1.50 g sample of solid NH₄NO₃ was added to 35.0 mL of water in a styrofoam cup (insulated from the environment) and stirred until it dissolved. The temperature of the solution dropped from 22.7°C to 19.4°C. What is the heat of reaction for dissolving NH₄NO₃, expressed in kJ/mol?

Answer 1

Answer : The heat of the reaction is, 1.27 kJ/mole

Explanation :

First we have to calculate the heat released.

Formula used :

`Q=m* c* \Delta T`

or,

`Q=m* c* (T_2-T_1)`

where,

Q = heat = ?

m = mass of sample = 1.50 g

c = specific heat of water =
`4.81J/g^oC`

`T_1` = initial temperature =
`22.7^oC`

`T_2` = final temperature =
`19.4^oC`

Now put all the given value in the above formula, we get:

`Q=1.50g* 4.81J/g^oC* (19.4-22.7)^oC`

`Q=-23.8095J=-0.0238kJ`

Now we have to calculate the heat of the reaction in kJ/mol.

`\Delta H=(Q)/(n)`

where,

`\Delta H` = enthalpy change = ?

Q = heat released = 0.0238 kJ

n = number of moles NH₄NO₃ =
`\frac{\text{Mass of }NH_4NO_3}{\text{Molar mass of }NH_4NO_3}=(1.50g)/(80g/mol)=0.01875mole`

`\Delta H=(0.0238kJ)/(0.01875mole)=1.27kJ/mole`

Therefore, the heat of the reaction is, 1.27 kJ/mole