answer 3. The stu dent then determined H tznfor the reaction of sodium hydroxide
ID: 547299 • Letter: A
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answer 3. The stu dent then determined H tznfor the reaction of sodium hydroxide and acetic acid, using the 00.0 mL of 0.8500M NaOH to 100.0 mL of procedure described in this module. The student added 1 b.8404M acetic acid. Prior to and following the mixing of the acid and base solutions, the following temperature-time data were collected. time min temperature, °C NaOH time, temperature, C min HOAc mixture 17.52 17.54 17.55 17.57 17.58 0.5 1.0 1.5 2.0 23.28 23.03 23.10 23.06 22.99 17.73 17.75 10 22.93 12 13 14 15 16 22.89 22.85 3.0 3.5 4.0 4.5 5.0 7.78 22.78 22.73 22.70 22.64 22.59 17.80 mixing 18 Plot the temperature-time data and determine the mean temperature of the unmixed reaExplanation / Answer
Ans. 7 : Limiting reagent is the reactant whose lesser equivalents are present in the reaction mixture, thus the reaction stops on the exhaustion of the limiting reagent. Here acetic acid is the limiting reagent as only 0.08404 moles of acetic acid are present where as 0.08500 moles of sodium hydroxide is present.
Ans.8 : dHneutralisation = m*c*dT
m= total mass or volume of the solution ( density of water = 1 g per centimeter cube ; here m=200)
c= specific heat capacity of the solution ( c=4.2 Joules per gram per Kelvin)
dT= change in temperature on mixing ( here 5.59oC )
Therefore ,
dHneutralisation= -(200*4.2*5.59) Joules ( negative sign signifies the release in energy)
= - (200*4.2*5.59)/(0.08404*1000) KJ per mole
= - 55.87 KJ per mole
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