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

Answer:

-15.24 kJ/mole of LiOH

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 =  $10 g LiOH + 120 g H_{2}O= 130 g solution$

Specific heat capacity of the solution =  $\frac{4.20 J}{g°C}$

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

Calculations:

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

$ΔH_{sln}=\frac{q_{sln}}{mole LiOH}$  Equation 1

$q_{sln}=-q_{calorimeter}=-mCΔT$  Equation 2  

$mole of LiOH=\frac{mas of LiOH}{Molecular weight of LiOH} Equation 3 Where:[tex]ΔH_{sln}$  = 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:

$moleLiOH=\frac{mass LiOH}{Molecular weight LiOH} =\frac{10 g LiOH}{23.96 g/mole} =0.4174 mole LiOH$

$ΔH_{sln}=\frac{-130 g solution*\frac{0.00420 kJ}{g solution°C}*(34.65-23) °C}{0.4174 moleLiOH}=-15.2393 \frac{kJ}{mole LiOH}$

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