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 :

$Q=m\times c\times \Delta T$

or,

$Q=m\times c\times (T_2-T_1)$

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:

$Q=1.50g\times 4.81J/g^oC\times (19.4-22.7)^oC$

$Q=-23.8095J=-0.0238kJ$

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

$\Delta H=\frac{Q}{n}$

where,

$\Delta H$ = enthalpy change = ?

Q = heat released = 0.0238 kJ

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

$\Delta H=\frac{0.0238kJ}{0.01875mole}=1.27kJ/mole$

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