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 = [tex]10 g LiOH + 120 g H_{2}O= 130 g solution[/tex]
Specific heat capacity of the solution = [tex]\frac{4.20 J}{g°C}[/tex]
Converting specific heat capacity to kJ/(g°C) =[tex][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][/tex]
Calculations:
To determine the dissolution enthalpy of LiOH, we can use the following equations:
[tex]ΔH_{sln}=\frac{q_{sln}}{mole LiOH}[/tex] Equation 1
[tex]q_{sln}=-q_{calorimeter}=-mCΔT[/tex] Equation 2
[tex]mole of LiOH=\frac{mas of LiOH}{Molecular weight of LiOH} Equation 3
Where:
[tex]ΔH_{sln}[/tex] = enthalpy of dissolution per mole of LiOH (kJ/mole).
[tex]q_{sln}[/tex] = heat released by dissolution (kJ).
[tex]q_{calorimeter}[/tex]=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:
[tex]moleLiOH=\frac{mass LiOH}{Molecular weight LiOH} =\frac{10 g LiOH}{23.96 g/mole} =0.4174 mole LiOH[/tex]
[tex]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}[/tex]
Note: Usually exothermic reactions like LiOH dissolutions, that release heat, results in negative enthalpy.
