Question

Consider the dissolution of ammonium nitrate. 1.25 g of amminum nitrate is dissolved in enough water to make 25.0 mL of solution. The initial temperature is 25.8 degC and the final temperature (after the solid dissolves) is 21.9 degC. Calculate the change in enthalpy for the reaction in kJ. Use 1.0 g/mL as the density of solution and 4.18 J/g*degC as the specific heat capacity.

Answer 1

Answer:

$\Delta H=-0.02\ kJ$

Explanation:

The expression for the calculation of the enthalpy change of a process is shown below as:-

$\Delta H=m\times C\times \Delta T$

Where,

$\Delta H$  is the enthalpy change

m is the mass

C is the specific heat capacity

$\Delta T$  is the temperature change

Thus, given that:-

Mass of ammonium nitrate = 1.25 g

Specific heat = 4.18 J/g°C

$\Delta T=21.9-25.8\ ^0C=-3.9\ ^0C$

So,

$\Delta H=-1.25\times 4.18\times 3.9\ J=-20.3775\ J$

Negative sign signifies loss of heat.

Also, 1 J = 0.001 kJ

So,

$\Delta H=-0.02\ kJ$