Question

40.0 mL of 1.00 M KOH(aq) and 40.0 mL of 0.500 M H2SO4(aq) are mixed in a coffee-cup calorimeter, both at 21.00°C. The final solution has a density of 1.02 g/mL, a volume of 80.0 mL, a specific heat of 4.00 J g-1 °C -1 , and a temperature of 27.85°C. Calculate the enthalpy change of this reaction per mole of H2O formed.

Answer 1

-56.4 kJ/mol

Step-by-step explanation:

There are two heat flows in this experiment.

Heat released by reaction + heat absorbed by solution = 0

q1 + q2 = 0

nΔH + mCΔT = 0

1. Moles of water formed

KOHL + H2SO4 → K2SO4 + 2H2O

V/mL: 40.0 40.0

c/(mol/L) 1.00 0.500

Moles of KOH = 40.0 mL × (1.00 mmoL/1 mL) = 40.00 mmol KOH

Moles of H2SO4 = 40.0 mL × (0.500 mmoL/1 mL) = 20.00 mmol H2SO4

Moles of H2O from KOH

= 40.00 mL KOH × (2.00 mmol H2O/2 mmoL KOH) = 40.00 mmol H2O

Moles H2O from of H2SO4

= 20.00 mL × (2 mmoL H2O/1 mmol H2SO4) = 40.00 mmol H2O

KOH and H2SO4 are present in equimolar amounts.

They form 40.00 mmol = 0.040 00 mol of water.

2. Calculate q1

q1 = 0.04000 mol × ΔH = 0.040 00ΔH J

3. Calculate q2

Mass of water formed = 0.040 00 mol × (18.02 g/1 mol) = 0.7208 g

V(solution) = 40.00 + 40.00 = 80.00 mL

Mass of solution = 80.00 mL × (1.02 g/1 mL) = 81.60 g

Total mass = 81.60 + 0.7208 = 82.32 g

ΔT = T2 - T1 = 27.85 - 21.00 = 6.85 °C

q2 = 82.32 × 4.00 × 6.85 = 2256 J

4. Calculate ΔH

0.040 00ΔH + 2256 = 0

0.040 00ΔH = -2256

ΔH = -2256/0.040 00 = = -56 400 J/mol = -56.4 kJ/mol

The enthalpy of neutralization is -56.4 kJ/mol.