When 28.0 mL of 0.515 M H 2 SO 4 is added to 28.0 mL of 1.03 M KOH in a coffee-c

Question

When 28.0 mL of 0.515 M H2SO4 is added to 28.0 mL of 1.03 M KOH in a coffee-cup calorimeter at 23.50

When 28.0 mL of 0.515 M H2SO4 is added to 28.0 mL of 1.03 M KOH in a coffee-cup calorimeter at 23.50 degree C, the temperature rises to 30.17 degree C. Calculate delta H of this reaction per mole of H2SO4 and KOH reacted. (Assume that the total volume is the sum of the individual volumes and that the density and specific heat capacity of the solution are the same as for pure water: d = 1.00 g/mL and c = 4.184 J/g times K.)

Explanation / Answer

Total Volume of the reaction mixture = 28 mL + 28 mL = 56 mL = 0.056 L

Mass of the mixture = density * volume = 1000*0.056/1000 = 0.056 kg

Change in temperature = 6.67 degree Kelvin

Heat released in the reaction = m*c*delta-T = 0.056*1000*4.184*6.67 = 1562.81 J

No. of moles of H2SO4 reacted = 0.028*0.515 = 0.01442 moles

No. of moles of KOH reacted = 0.028*1.03 = 0.02884 moles

Hence, delta-H = 1562.81 / 0.01442 = 108377.95 J per mole of H2SO4

delta-H = 1562.81 / 0.02884 = 54188.97 J per mole of KOH

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