When 25.5 mL of 0.500 M H2SO4 is added ot 25.5 mL of 1.00 M KOH in a coffee-cup

Question

When 25.5 mL of 0.500 M H2SO4 is added ot 25.5 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50 degrees Celsius, the temmperature rises to 30.17 degrees Celsius. Calculate delta H of this reaction. (Assume that the total volume is the sum of the individual volumes and the density and specific heat capacity of the solution are the same as for pure water) (d for water=1.00g/mL; c for water=4.184 J/g degrees Celsius)

Explanation / Answer

When 31.5 mL of 0.520 M H2SO4 is added to 31.5 mL of 1.04 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ?H of this reaction. (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.) Calculate: ?H per mole of H2SO4 reacted ?H per mole of KOH reacted First write down the balanced chemical reaction: H2SO4 + 2 KOH -> K2SO4 + H2O Then calculate how many moles of each reactant you have: moles H2SO4 = (0.0315L)*(0.520M) = 0.01638 moles H2SO4 moles KOH = (0.0315L)*(1.04M) = 0.03276 moles KOH From this information determine which reactant, if any, runs out before the other: Since it requires 2 moles of KOH for every mole of H2SO4 you see that both will be used up in the reaction. Next, calculate the energy released from: E=mc(T2-T2) = (31.5ml + 31.5ml)*(1g/ml)*(1cal/gC)*(30.17C-23.50C… = 420.2 cal Now simply calculate the cal/mole for each reactant: deltaH = (420.2cal)/(0.01638 mole H2SO4) = 6.88 cal/mole H2SO4 deltaH = (420.2cal)/(0.03276 mole KOH) = 3.44 cal.mol KOH You can compare these results with your answer. It might also be possible there is a "typo" in the book's answer. Hope this helps.

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