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In: Chemistry

When 24.7 mL of 0.500 M H2SO4 is added to 24.7 mL of 1.00 M KOH...

When 24.7 mL of 0.500 M H2SO4 is added to 24.7 mL of 1.00 M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C. Calculate ΔH 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.) (d for water = 1.00 g/mL; c for water = 4.184 J/g·°C.)

.................. kJ/mol H2O

Expert Solution

moles of acid H2SO4 = 24.7 x 0.50 / 1000 = 0.0124

moles of NaOH = 24.7 x 1.0 / 1000 = 0.0247

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

1 2

0.0124 0.0247

here both are consumed

moles of water fomred = 0.0247

total volume = 24.7 + 24.7 = 49.4 mL

mass of solution = volume x density

= 49.4 x 1

= 49.4 g

dT = 30.17 -23.50 = 6.67 oC

Q = 49.4 x 4.184 x 6.67

Q = 1379 J

molar enthalpy of neutralization = - Q / n

= -1.379 / 0.0247

= - 55.8 kJ / mol

queen_honey_blossom answered 2 years ago

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