>>>Heat of Reaction
Aim: To determine the heat of reaction of NaOH with HCl
Apparatus\Materials: Polythene Cup,
Thermometer,
Measuring Cylinder,
NaOH,
HCl.
Procedure: (1) 25cm" of HCl was placed in polythene cup,
(2) HCl was left to stand for 5 minutes and then the temperature was taken
(3) Steps 1-2 was repeat with NaOH instead of HCl,
(4) NaOH was added to HCl in polythene cup and stirred for 5minutes and the highest temperature the liquid rose to was recorded.
Results: Initial temperature of NaOH - 32?
Initial temperature of HCl - 31?
Final temperature of mixture - 38?
Temperature rise for NaOH = 6?
Temperature rise for HCl = 7?
Volume of mixture = 50cm"
Mass of solution = 50g
1 cm" of solution has a mass of 1g
Number of moles present in 25cm" of NaOH = .0025moles
Heat produced = 1365KJ
Calculations: (1) Mass of 1cm of solution = 1g since the density of the solution is 1g per cm",
(2) Number of moles of NaOH
= volume incm"xconcentration in moles per dm"
1000 cm"
= (25 x .1)/1000
= .0025moles
(3) Heat produced = MC?? of NaOH+ MC?? of HCl
= .025kg X 4200 X (38?- 32?) NaOH
= 630J (NaOH)
= 025kg X 4200 X (38?- 31?) HCl
= 735J (HCl)
= 630J (NaOH) + 735J (HCl)
= 1365J
Conclusion: An exothermic reaction takes place when NaOH is mixed with HCl. The amount of heat produced when .0025 moles of NaOH is mixed which .0025 Moles of HCl is equal to 1365J and the amount of heat that will be produced when 1mole of NaOH reacts with 1mole of HCL will be equal to ((1 moles\ .1mole) X 1356j), 546KJ
Good lab
Good lab.
Well organized.
Good conclusion.
And excellent calculations!!
Needs sources
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