Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

In a constant-pressure calorimeter, 65.0 mL of 0.870 M H2SO4 was added to 65.0 m

ID: 538432 • Letter: I

Question

In a constant-pressure calorimeter, 65.0 mL of 0.870 M H2SO4 was added to 65.0 mL of 0.480 M NaOH. The reaction caused the temperature of the solution to rise from 21.11 °C to 24.38 °C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184 J/g·K, respectively), what is H for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes.
Map Map Sapling Learning macmilan learning In a constant-pressure calorimeter, 65.0 mL of 0.870 M H2S04 was added to 65.0 mL of 0.480 M NaOH The reaction caused the temperature of the solution to rise from 21.11 to 24.38 . If the solution has the same density and specific heat as water (1.00 g/mL and 4.184 J/g. K, respectively), what is dH for this reaction (per mole of HzO produced)? Assume that the total volume is the sum of the individual volumes. Number kJ/ mol H,O

Explanation / Answer

H2SO4 + 2NaOH ---> Na2SO4 + 2H2O

limiting reactant

moles H2SO4 = 0.870 M x 65 ml = 56.55 mmol

moles NaOH = 0.480 M x 65 ml = 31.2 mmol

NaOH is the limiting reactant

moles of H2O produced = 31.2/2 x 1000 = 0.0156 mol

dT = (24.38 - 21.11) oC

mass of solution (m) = 65 + 65 = 130 g

So,

dH = mCpdT/moles H2O

      = 130 x 4.184 (24.38 - 21.11)/0.0156

      = 114.014 kJ/mol