Question

Part A: Heat Capacity of Calorimeter:

 

 1. Heat Capacity of the Calorimeter: 

 2. Heat of Reaction:

 3. Moles of Product Produced:

 4. Molar Enthalpy of Neutralization

 Part C: Molar Enthalpy of Solution for Trial 1:

 6. Heat of Reaction:

 7. Moles of NaHCO3 dissolved:

 8. Molar Enthalpy of solution:

 

Answer 1

Part- A-

Given we have taken calorimeter with cold water with initial temp (22.1oC) and after addition of hot water, then final temp of all becomes (35oC)

That means heat is transfered from hot to cold water untill all of them come to equal temp i.e 35oC

Now heat lost by hot water (-Qhot) = Heat gained by cold water (Qcold) + Heat gained by calorimeter (Qcal)

Here heat lost will bear a negative sign where as the heat gained will have a positive sign.

Now formula to calculate heat lost by hot water (-Qhot) is-

(-Qhot) = m * C * del T

where

m = mass of hot water taken

= 112.5 g - 65.4 g

= 47.1‬ g

C = Heat capasity of water

= 4.18 J/goC

del T = Temp change for hot water

= final temp - initial temp

= 35oC - 49.5oC

= -14.5oC

So putting the values-

(-Qhot) = m * C * del T

= 47.1‬ g * 4.18 J/goC * -14.5oC

= -2,854.7 J

Similalrly

The formula to calculate heat gained by cold water (Qcold) is-

(Qcold) = m * C * del T

where

m = mass of cold water taken

= 65.4 g - 18.7 g

= 46.7‬ g

C = Heat capasity of water

= 4.18 J/goC

del T = Temp change for cold water

= final temp - initial temp

= 35oC - 22.1oC

= 12.9‬oC

So putting the values-

(-Qcold = m * C * del T

= 46.7‬ g * 4.18 J/goC * 12.9‬oC

= 2,518.15 J

That means now we have-

heat lost by hot water (-Qhot) = Heat gained by cold water (Qcold) + Heat gained by calorimeter (Qcal)

2,854.7 J = 2,518.15 J + Heat gained by calorimeter (Qcal)

Heat gained by calorimeter (Qcal) = 2,854.7 J - 2,518.15 J

= 336.55‬ J

Again The formula to calculate heat gained by calorimeter (Qcal) is-

(Qcal) = C * del T

where

C = Heat capasity of calorimeter

= ?

del T = Temp change for calorimeter

= same as that for cold water

= 12.9‬oC

So putting the values-

(Qcal = C * del T

336.55‬ J = C * 12.9‬oC

C = 336.55‬ J / 12.9‬oC

= 26.09 J/oC

Part- B-

Trial 1

Here neutralization reaction takes place between HCl and NaOH as a result of which, the final temp of both the solution as well as calorimeter increases.

So heat lost by reactionr (-Qrxn) = Heat gained by solution (Qsol) + Heat gained by calorimeter (Qcal)

Here heat lost will bear a negative sign where as the heat gained will have a positive sign.

Now

The formula to calculate heat gained by solution (Qsol) is-

(Qsol) = m * C * del T

where

m = mass of solution

= mass of HCl + NaOH

= (26.9 - 14.4) + (40.4 - 14.4)

= 38.5‬ g

C = Heat capasity of solution = Heat capasity of water

= 4.18 J/goC

del T = Temp change solution

= final temp - initial temp

= 25.6oC - average einitial temp of HCl and NaOH

= 25.6oC - (20.25oC‬)

= 5.35oC

So putting the values-

(Qsol) = m * C * del T

= 38.5 g * 4.18 J/goC * 5.35‬oC

= 860.98 J

Again The formula to calculate heat gained by calorimeter (Qcal) is-

(Qcal) = C * del T

where

C = Heat capasity of calorimeter

= 26.09 J/oC

del T = Temp change for calorimeter

= 5.35‬oC

So putting the values-

(Qcal = C * del T

= 26.09 J/oC * 5.35‬‬oC

= 139.58 J

So putting the values-

heat lost by reactionr (-Qrxn) = Heat gained by solution (Qsol) + Heat gained by calorimeter (Qcal)

= 860.98 J + 139.58 J

= 1,000.56‬ J

Now we have mols of HCl = mass of HCl taken / molar mass of HCl

= (26.9 - 14.4)g / 36.46 g/mol

= 12.5‬ g / 36.46 g/mol

= 0.34 moles

And we have mols of NaOH = mass of NaOH taken / molar mass of NaOH

= (40.4 - 14.4)g / 39.997 g/mol

= 26‬‬ g / 39.997 g/mol

= 0.65 moles

Now the reaction between HCl and NaOH is-

HCl + NaOH ----------------> NaCl + H2O

i.e 1 mole of HCl require 1 mole of NaOH for complete neutralization.

So from our mols of HCl and NaOH calculated, we can see HCl is the limiting reagent = 0.34 moles

So

Enthlpy of reaction = heat lost by reaction / mols sof limiting reagent

= -1,000.56‬ J / 0.34 moles

= -2943 kJ/mol

3-

And mols of product formed is as per the limiting reagent i.e 0.34 moles of NaCl