1. A \"coffee-cup\" calorimetry experiment is run for the dissolution of 2.5 g o
ID: 1033978 • 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. 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?
There are few assumptions that need to be made before the system and sorroundings are considered. The assumption being that there is no loss of heat from the caloriemeter to the environment, only in which case we can restrict on the sorrundings. So with that assumption, the caloriemeter and all the water that had same temp as caloriemeter initially would be included in the sorroundings. All the other components contained in the caloriemeter would be counted towards the system. More specifially, the reactants and products would be considered as the system.
2.Continuing with the same expt as in Q 1, is the system exothermic or endothermic? How can you tell?
The system is exothermic as the temperature has gone up from 23.5 to 28.3 indicating that the heat was given out to the sorroundings.
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?
Total mass of the solution = mass of lithium nitrate + mass of water
mass of lithium nitrate = 2.5 g
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.
mass of lithium nitrate = 99.7 g
mass of water = Volume of water * density of water = 97.2 ml * 1.0 g/ml = 97.2 g
Total mass of the solution = 2.5 g + 97.2 g = 99.7 g
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.
Since the temperature os the water incresed, water being sorroundings has gained/absorbed heat.
The heat absorbed by the sorroundings = Total mass of soultion* Specific heat of water * change in temperature
= 99.7 g * 4.184 J/g0C * (28.3 -23.50C)
= 2000.3 J = 2.00 KJ
Since the heat was absorbed by the sorroundings, the sign would remain +ve.
5. How many moles of lithium nitrate are used in this experiment?
Moles of lithium nitrate used = mass of lithium nitrate/ molar mass of lithium nitrate
= 2.5 g / 68.95 g/mol = 0.0363 moles lithium nitrate
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.
Lithium nitrate is part of the system, so heat is lost by the system as it is gained by the sorroundings.
So heat lost by system = -2000.3 J. This is the heat lost by 0.0363 moles of lithium nitrate, as that was the number of moles of lithium nitrate consumed.
=> Enthalpy of 1 mole of lithium nitrate = heat lost/ moles lithium nitrate = -2.00 KJ /0.0363 moles = -55.16 KJ/mol
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