When 26.0 ml of 0.500 m h2so4 is added to 26.0 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.)

The ΔH of this reaction is 55.8 kJ/mol.

Explanation:

Moles of sulfuric acid:

Moles of sulfuric acid =

Moles of Potassium hydroxide:

Moles of Potassium hydroxide=

1 mol of sulfuric acid reacts witrh 2 mol of potassium hydroxide.

Then 0.013 mol of  sulfuric acid will react with:

of potassium hydroxide.

Total volume of the solution = 26.0 mL+26.0 mL= 52 .0 mL

The density of the solution is same as  pure water = 1.00g/mL (given)

Mass of the solution ,m=

The specific heat capacity of the solution is same as  pure water:

c =4.184 J/g°C(given)

Change in temperature of the solution = ΔT =30.17°C - 23.50°C=6.67°C

Heat released during the mixing of  both the solution; Q

Q  =1,451.17 J = 1.4511 kJ

When 0.013 mol of sulfuric acid reacts with 0.026 moles of potassium to give 0.026 moles of water.

1.4511 kJ of heat is released when 0.026 moles of water are formed.

Then , for 1 mole of water the energy release will be:

So, the ΔH of this reaction is 55.8 kJ/mol.