Question

In a coffee cup calorimeter, 1.60 g of NH4NO3 is mixed with 75.0 g of water at an initial temperature of 25.008C. After dissolution of the salt, the final temperature of the calorimeter contents is 23.348C. Assuming the solution has a heat capacity of 4.18 J 8C21 g21 and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of NH4NO3 in units of kJ/mol.

Answer 1

Answer: ΔH for the dissolution of
`NH_4NO_3` is +26.0205 kJ/mol

Step-by-step explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

`Q=m* C* \Delta T`

Q = Heat released by solution = ?

C = heat capacity =
`4.18J/g^0C`

Initial temperature of water =
`T_i` =
`25.008^0C`

Final temperature of water =
`T_f` =
`23.348^0C`

Change in temperature ,
`\Delta T=T_f-T_i=(23.348-25.008)^0C=-1.66^0C`

Putting in the values, we get:

`Q=75.0g* 4.18J/g^0C* -1.66^0C=-520.41 J`

As heat released by water is equal to heat absorbed by dissolution of
`NH_4NO_3`

`\text{Moles of}NH_4NO_3=\frac{\text{given mass}}{\text{Molar Mass}}=(1.60g)/(80g/mol)=0.02mol`

Enthalpy change for 0.02 moles of
`NH_4NO_3` = 520.41 J

Enthalpy change for 1 mole of
`NH_4NO_3` =
`(520.41)/(0.02)* 1=+26020.5J=+26.0205kJ`

ΔH for the dissolution of
`NH_4NO_3` is +26.0205 kJ/mol