Question

This question has multiple parts.

a. In a coffee-cup calorimeter, 1.81 g NH4NO3 was mixed with 85.00 g water at an initial temperature 25.00oC. After dissolution of the salt, the final temperature of the calorimeter contents was 23.34oC.
Assuming the solution has a heat capacity of 4.18 J/g∙oC, and assuming no heat loss to the calorimeter, calculate the enthalpy of solution (∆Hsoln) for the dissolution of NH4NO3 in units of kJ/mol.
∆Hsoln___ = kJ/mol
b. In a coffee-cup calorimeter, 1.81 g NH4NO3 was mixed with 85.00 g water at an initial temperature 25.00oC. After dissolution of the salt, the final temperature of the calorimeter contents was 23.34oC.
If the enthalpy of hydration for NH4NO3 is -630. kJ/mol, calculate the lattice energy of NH4NO3.
Lattice energy ___= kJ/mol

Answer 1

Δ
`H_(solution)` = 26.20 kJ/mol

Lattice Energy ( - Δ
`H_1` ) = - 656.20 kJ/mol

Step-by-step explanation:

Given that:

the mass of weight of solute present = 1.81 g

number of moles of NH₄NO₃ =
`(mass of weight )/(molar mass)`

number of moles of NH₄NO₃ =
`(1.81 g)/(80.0g/mol)`

number of moles of NH₄NO₃ = 0.023 mol

The equation for the dissolution of NH₄NO₃ can be written as:

`NH_4NO_(3(aq)) -------> NH^+_(4(aq)) +NO_(3(aq))^-`

Since heat is absorbed by the salt ( NH₄NO₃) after dissolution of the salt in the water: we have;

Heat absorbed by the NH₄NO₃ = Heat evolved by the solution.

Let first determine the heat evolved by the solution in order to find the amount of heat absorbed by the NH₄NO₃.

Heat of solution is given as:

`q_(solution)= mS \delta T`

where m = total mass of the weight of the solution = 1.81 g + 85.00 g = 86.81 g

`\delta T` = 25.00°C - 23.34°C

`\delta T` = 1.66°C

S = specific heat capacity of the solution which is given as : 4.18 J/g °C

`q_(solution)= - mS \delta T` (since heat is lost by the water to the compound)

= -86.81 g × 4.18 J/g °C × 1.66°C

= - 602.357228 J

= - 602.36 J

This same amount of heat is absorbed by 1.81 g of NH₄NO₃ (0.023 mol)

Hence; The amount of heat absorbed by 1 mole of NH₄NO₃

=
`(amount of heat absorbed)/(number of molesin solution)`

=
`(602.36J)/(0.023)*(1kJ)/(1000J)`

= 26189.565 × 0.001

= 26.189565

≅ 26.20 kJ/mol

Hence, the Δ
`H_(solution)` = 26.20 kJ/mol

b)

The heat of solution Δ
`H_(solution)` = Δ
`H_(hyd)` + Δ
`H_1`

where;

Δ
`H_(hyd)` = the enthalpy of hydration for NH₄NO₃ is -630. kJ/mol

Δ
`H_1` is said to be the energy needed for dissociation of the NH₄NO₃ in the solution.

∴ from the above equation;

Δ
`H_1` = Δ
`H_(solution)` - Δ
`H_(hyd)`

Δ
`H_1` = 26.20 kJ/mol - (- 630 kJ/mol)

Δ
`H_1` = 26.20 kJ/mol + 630 kJ/mol

Δ
`H_1` = 656.20 kJ/mol

However, Lattice Energy = - Δ
`H_1`

∴ Lattice Energy = - 656.20 kJ/mol

Answer 2

A)
`\Delta H_(sol)=26.653 kJ/mol`

Step-by-step explanation:

Hi,

A) First we need to calculate the heat consumed (because the temperature decreased) by the dissolution process.

All the heat consumed was shown in the change of temperature given that there is no heat loss in the calorimeter.

So:

`Q=Cp*m*\Delta T`

We know that:

`Cp=4.18 (J)/(g*C)`

`\Delta T=25.00 C - 23.34 C= 1.66 C`

`m=m_(water) + m_{salt]=1.81g +85g= 86.81g`

The heat is:

`Q=4.18 (J)/(g*C)*1.66 C*86.81g`

`Q=602.36J`

1 mol of the amonium nitrate weighs: 80g/mol:

`n=(1.81g)/(80g/mol)`

`n=0.0226mol`

The heat for 1 mol:

`\Delta H_(sol)=(1)/(0.0226) * 602.36J`

`\Delta H_(sol)=26653 J/mol=26.653 kJ/mol`