2. A student constructs a coffee cup calorimeter and places 50.0 mL of water int

Question

2. A student constructs a coffee cup calorimeter and places 50.0 mL of water into it. After a brief period of stabilization, the temperature of the water in the calorimeter is determined to be 23.50°C. To this is added 50.0 mL ofwater that was originally at 47.00 A careful plot of the temperatures recorded after this established that the temperature at T was 34.25 C. What is the calorimeter constant in J/°C for this calorimeter? (density HO - 1.00 g/mL; specific heat of water 4.184 J/g. °O) 3. 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 22.0°C. The final temperature of the neutralization reaction was determined to be 33.5 °C. The calorimeter constant was known to be 20.6 J/ (specific heat of H20-4.184 J/ g-C). Show all work a. What is the total amount of heat evolved in this reaction? b. If 135 mmol of the monoprotic acid were neutralized in this reaction, what is the molar heat of neutralization for this reaction?

Explanation / Answer

2) calorimeter constant = C

    heat absorbed by calorimeter = heat lost by hot water - heat gained by cold water

       C*DT = m2*S*DT - m1*s*DT

      C*(34.25-23.5) = 50*4.184*(47-34.25)-50*4.184*(34.25-23.5)

    C = 38.921 j/c

3) heat evolved in the reaction(q) = m*s*DT + C*DT

   m = mass of mixture = 50+50 = 100 g

   s = specific heat of solution = 4.184 j/g.c

   DT = 33.5-22 = 11.5 c

   C = calorimeter constant = 20.6 j/c

q = 100*4.184*11.5+20.6*11.5

     = -5.05 kj

DHrxn = q/n

   n = no of mol of monoprotic acid = 0.135 mol

        = -5.05/0.135 = -37.4 kj/mol

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