When 30.5 mL of 0.580 M H2SO4 is added to 30.5 mL of 1.16 M KOH in a coffee-cup calorimeter at 23.50°C, the te?
When 30.5 mL of 0.580 M H2SO4 is added to 30.5 mL of 1.16 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate 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×K.)
H per mole of H2SO4 reacted: (kJ/mol)
H per mole of KOH reacted: (kJ/mol)
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4.184 J/g x K tells us the amount of heat produced for every rise in degree K.
The temperature rise was 30.17 – 23.5 = 6.67C, which is also 6.67K, since C and K have the same scale, with different starting points.
6.67 * 4.184 = 27.90728 J is the amount of heat produced.
Now convert it to “per mole” of each ingredient, because this is per “30.5ml of .58M H2S04”.
The reaction is:
H2S04 + 2KOH -> K2S04 + 2H2O
The .58M HS204 means .58 moles HS204 per liter.
Since we have 30.5 ml of .58M H2S04, we have 30.5ml/1000 ml/L * .58moles/L = .01769 Moles of H2S04
For the KOH: we have 30.5/1000 * 1.16 = .026293 Moles of KOH
So, how much reacted? We either had too much KOH or too much H2SO4.
For 0.026293 Moles of KOH, we should have 0.01314 moles of H2S04 to react completely.
That means we had excess H2S04 (.01769 Moles instead of the .01314 Moles needed to react).
So, the per mole of KOH is our heat divided by our Moles: 27.90728 J/0.026293 = 1061.395809 J/mole KOH
Since only half the number of moles of HS204 was used in the reaction, half that is the heat per mole of H2S04, or 1061.395809/2 = 530.6979044 J/mole HS204.