Question

1. When 185 mL of 0.431 M HCl at 20.48 ^oC is mixed with 215 mL of 0.431 M NaOH at 20.48 ^oC in a coffee cup calorimeter, the temperature of the mixture rises to 27.8 ^oC. What is the heat of reaction in KJ per mole of the acid? Assume the density and the specific heat of the solution are 1.02 g/ml and 4.017 J/g.^oC respectively. The heat capacity of the calorimeter is 10.0 J/^oC.

Answer 1

Here after the reaction of HCl and NaOH the final temperature of the solution is increasing. That means heat is released by the reaction which is completely absorbed by both the solution and calorimeter to raise their temperatures.

So

heat released from the reaction (Qrxn) = Heat gained by the solution(Qsol) + Heat gained by calorimeter (Qcal)

Here the heat lost will be negative is sign where as the heat gains are positive.

Now formula to calculate the Heat gained by the solution(Qsol) is

(Qsol) = m * C * del T

where

m = mass of solution

= density * volume of solution

= 1.02 g/ml * (185 mL + 215 mL)

= 1.02 g/ml * (400‬ mL)

= 408‬ g

C = Heat capasity of solution

= 4.017 J/g.oC

del T = Change in temp

= final temp - Initial temp

= 27.8 oC - 20.48 oC

= 7.32‬oC

So putting these values-

(Qsol) = m * C * del T

= 408‬ g * 4.017 J/g.oC * 7.32‬oC

= 11,997 J

= 11.997 kJ

Similarly formula to calculate the Heat gained by the calorimeter(Qcal) is

(Qcal) = C * del T

where

C = Heat capasity of calorimeter

= 10.0 J/g.oC

del T = Change in temp   

= 7.32‬oC

So putting these values-

(Qcal) = C * del T

= 10.0 J/g.oC * 7.32‬oC

= 73.2‬ J

= 0.0732 kJ

So now

heat released from the reaction (Qrxn) = Heat gained by the solution(Qsol) + Heat gained by calorimeter

= 11.997 kJ + 0.0732 kJ

= 12.07 kJ

Since it is negative in sign, we can write (Qrxn) = -12.07 kJ

Now the Enthalpy of the reaction per mole of the acid = heat released from the reaction (Qrxn) / moles of acid

= (Qrxn) / [ concentration * volume]

= (Qrxn) / [ 0.431 M * 185 mL]

= (Qrxn) / [ 0.431 mol/1000 mL * 185 mL]

= (-12.07 kJ) / [ 0.0797 mols]

= -151.4 kJ/mol