Answer to Question #188936 in General Chemistry for rainiel mayuyo

Question

 A student transferred 50.0 mL 1.00 M HCl into a coffee-cup calorimeter, which had a temperature of 25.5 oC. He then added 50.0 mL 1.00 M NaOH, which also had a temperature of 25.5 oC, and stirred the mixture quickly. The resulting solution was found to have a temperature of 32.5 oC. The calorimeter constant for the coffee-cup calorimeter used was 15.0 J/ oC. Calculate the heat of reaction.

Answer 1

The Reaction undergoing is

HCl (aq) + NaOH (aq)$\implies$ NaCl(aq) +  H₂O (l)

Volume Of HCl Solution = 50 ml 

Volume Of NaOH Solution = 50 ml

Total Volume of Solution = (50 +50) ml

                    = 100 ml

Density of water = 1g/ml

Mass of solution = 100 ml × 1g/ml

                     = 100 g

Temperature Before mixing = T_b = 25.5°C

Temperature after mixing = T_f = 32.5°C

Change in Temperature = ∆T

∆T = T_f - T_b 

     = (32.5- 25.5)°C

            = 7°C

Specific heat capacity of water = c= 4.18 J/g°C

Heat gained by Solution = q_s = mc∆T

  q_s = 100g × 4.18J/g°C×7°C

              q_s = 2926 J

calorimeter constant =Heat Capacity of calorimeter = Heat gained by calorimeter

Temprature change

Heat gained by calorimeterTemprature change

Heat gained by calorimeter qc = Calorimeter constant ×Temperature change

qc = 15.0 J/°C  ×7 °C

     qc = 105 J

qreaction + qSolution + qcalorimeter = 0

Heat of reaction = -( Heat gained by Solution( qs )+ Heat gained by calorimeter( qc )

Q=2926 J+105 J=3033.8J

Moles of HCl=$\frac{M×V}{1000}$

$=\frac{1.0×50}{1000}$

$=0.05moles$

$Heat of reaction=\frac{Q}{Moles}$

$=\frac{3033.8}{0.05}=60,676J$

1Joule=0.001Kilojoule

$\therefore60,676J=?$

$\frac{60,676J×0.001kJ}{1J}$

$=60.68kJ$