1. A \"coffee-cup\" calorimetry experiment is run for the dissolution of 2.5 g o
ID: 1033975 • Letter: 1
Question
1. A "coffee-cup" calorimetry experiment is run for the dissolution of 2.5 g of lithium nitrate placed into 97.2 mL of water. The temperature of the solution is initially at 23.5 oC. After the reaction takes place, the temperature of the solution is 28.3 oC.
For this process, what is the system and what is the surroundings?
2.Continuing with the same experiment as in question 1, is the system exothermic or endothermic? How can you tell?
3.Using a density of 1.0 g/mL for the water added and adding in the mass of the lithium nitrate, what is the total mass of the solution and solid?
4.How much heat was absorbed or lost by the surroundings? Use 4.184 J/goC for the specific heat of the solution. Put your answer in units of kJ and make sure the sign is correct.
5.How many moles of lithium nitrate are used in this experiment?
6.What would be the enthalpy for the dissolution reaction of one mole of lithium nitrate? Put your answer in kJ/mol and watch the sign for the enthalpy.
Explanation / Answer
1. system = solution containing coffee-cup calorimeter.
surroundings = rest of the universe.
2. as the temperature of rises, means energy released.
so that, it is exothermic process.
3. total mass of the solution and solid = 97.2*1+2.5 = 99.7 g
4. heat absorbed by solution(q) = m*s*DT
m = mass of solution = 99.7 g
s = specific heat of the solution = 4.184 j/g.c
DT = 28.3-23.5 = 4.8
q = 99.7*4.184*4.8
= 2002.3 joule
heat absorbed by surroundings = +2002.3 joule
5. no of mol of LiNO3 = W/mWT = 2.5/68.95 = 0.036 mol
6. DHsol = -q/n = -2.0/0.036 = -55.55 kj/mol