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.

-56.4 kJ/mol

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.

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o