Answer to Question #188936 in General Chemistry for rainiel mayuyo

Question #188936

 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.


1
Expert's answer
2021-05-04T14:15:56-0400

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 = Tb = 25.5°C

Temperature after mixing = Tf = 32.5°C

Change in Temperature = ∆T

∆T = Tf - Tb 

     = (32.5- 25.5)°C

            = 7°C

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

Heat gained by Solution = q= mc∆T

  qs = 100g × 4.18J/g°C×7°C

              qs = 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\u00d7V}{1000}"

"=\\frac{1.0\u00d750}{1000}"


"=0.05moles"

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


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


1Joule=0.001Kilojoule

"\\therefore60,676J=?"

"\\frac{60,676J\u00d70.001kJ}{1J}"


"=60.68kJ"



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Comments

Assignment Expert
05.05.21, 08:50

Dear Rainiel, You're welcome. We are glad to be helpful. If you liked our service please press like-button beside answer field. Thank you!

Rainiel
05.05.21, 01:29

Thank you so much!!

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