Question

6. You mix 50.0 mL of a weak monoprotic acid with 50.0 mL of NaOH solution in a coffee cup calorimeter. Both solutions (and the calorimeter) were initially at 22.0OC. The final temperature of the neutralization reaction was determined to be 22.5OC. a) What is the total amount of heat evolved in this reaction? Show all work. b) If 0.135 moles of the monoprotic acid were neutralized in this reaction, what is the molar heat of neutralization (enthalpy) for this reaction?

Answer 1

100gms (100 ml ). of solution gained (22.5 - 22.0 ) = 0.5 ^o C

So, heat change = mass of solution x specific heat capacity of water x rise in temperature

------------------------- = ( 100 ) ( 4.184 ) ( 0.5 )

------------------------ = 209. 2 joules

heat taken up by calorimeter = calorimeter constant * x rise in temperature

------------------------------------------ = 20.6 x 0.5

------------------------------------------ = 10.3 joules

[ The Calorimeter constant is known to be = 20.6 joules per degree Celcius ]

Hence total amount of heat evolved in this neutralization reaction = ( 209.2 + 10.3 )

---------------------------------------------------------------------------------------------- = 219 .5 joules

b )

moles of water formed in this neutralization reaction = moles of monoprotic acid given

--------------------------------------------------------------------------- = 0.135

Therefore molar heat of neutralization ( enthalpy ) for this reaction = heat evolved / number of moles of acid

--------------------------------------------------------------------------------------------------- = 119.5 / 0.135

-------------------------------------------------------------------------------------------------- = 1625.92 joules

-----------------------------------------------------------------------------------------------or, = 16.26 KJ mole^-1  

Note that specific heat of water = 4.184 joules per degree Celcius

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