Question

A 10.0 g sample of LiOH is dissolved in 120.00 g of water at 23.0°C in a coffee-cup calorimeter with a heat capacity of 100 J/°C. The heat capacity of the solution was 4.20 J/g°C, and final temperature was 34.65°C. Based on these measurements, what is the enthalpy of dissolution of LiOH in kJ/mol?

Answer 1

-15.24 kJ/mole of LiOH

Step-by-step explanation:

General considerations:

-A coffee-cup calorimeter is a device that works at constant pressure. Since enthalpy (ΔH) is defined as a heat flow at constant pressure, the coffee-cup calorimeter is usually used to measure enthalpy changes in processes at constant pressure.

-The high heat capacity of calorimeter indicates its difficulty to vary its temperature.

-The calorimeter absorbs a negligible amount of heat.

Information given in the statement:

Intial temperature = 23°C

Final temerature = 34.65°C

Mass of LiOH= 10 g LiOH

Mass of solution =
`10 g LiOH + 120 g H_(2)O= <strong>130 g solution</strong>`

Specific heat capacity of the solution =
`(4.20 J)/(g°C)`

Converting specific heat capacity to kJ/(g°C) =
`[tex](4.20 J)/(g°C)=(4.20 J)/(g°C)*(1 kJ)/(1000 J)=\<strong>frac{0.00420 kJ}{g°C}</strong>`[/tex]

Calculations:

To determine the dissolution enthalpy of LiOH, we can use the following equations:

`ΔH_(sln)=(q_(sln))/(mole  LiOH)` Equation 1

`q_(sln)=-q_(calorimeter)=-mCΔT` Equation 2

`mole of LiOH=(mas of LiOH)/(Molecular weight of LiOH) <em><strong>Equation 3</strong></em></p><p> </p><p>Where:</p><p>[tex]ΔH_(sln)` = enthalpy of dissolution per mole of LiOH (kJ/mole).

`q_(sln)` = heat released by dissolution (kJ).

`q_(calorimeter)`=heat absorbed by the solution in calorimeter (kJ) .

m=mass of solution (g).

C=specific heat capacity of the solution (kJ/g°C).

ΔT=chage of temperature of the solution in calorimeter, final temperature minus initial temperatrue (°C).

Molecular weight of LiOH=23.95 g/mole

Replacing the given data in equations 1, 2 and 3, we get:

`moleLiOH=(mass LiOH)/(Molecular  weight  LiOH) =(10 g LiOH)/(23.96 g/mole) =<strong>0.4174 mole LiOH</strong>`

`ΔH_(sln)=(-130 g solution*(0.00420 kJ)/(g solution°C)*(34.65-23) °C)/(0.4174 moleLiOH)=-<strong>15.2393</strong> (kJ)/(mole LiOH)`

Note: Usually exothermic reactions like LiOH dissolutions, that release heat, results in negative enthalpy.