When 25.0 ml of 0.500 m h2so4 is added to 25.0 ml of 1.00 m koh in a coffee-cup calorimeter at 23.50°c, the temperature rises to 30.17°c. calculate h of this reaction per mole of koh. (assume that the total volume is the sum of the volumes and that the density and specific heat capacity of the solution are the same as for water.) ?

The chemical reaction that occurs between the given substances is a neutralization reaction as shown below:

H_2SO_4 + 2KOH - > K_2SO_4 + 2H_2O

1 mol 2 mol 1 mol 2 mol

The number of moles of the given substances is calculated as shown:

Number of moles of H_2SO_4 = 25.0 mL x 0.50 M = 12.5 millimoles

Number of moles of KOH = 25.0 mL x 1.00 M = 25.0 millimoles

As 1 mol of sulfuric acid reacts with 2 mol of KOH to give 2 mol of water, 12.5 millimoles of sulfuric acid completely reacts with 25.0 millimoles of KOH to give 25.0 millimoles of water.

Total volume of the solution = 25.0 mL + 25.0 mL = 50.0 mL.

Density of water is 1 g/mL. Use this to calculate the mass of the solution.

Mass of the solution - 50.0 mL x 1 g/mL = 50.0

The specific heat of water is 4.184 J/gK. The temperature of the solution is increased from 23.5 degrees Celsius to 30.17 degrees Celsius.

The amount of heat released = 4.184 J/gK x 50.0 g x (30.17C - 23.50C) 1395 J

The amount of heat released per one mole of water formed can be calculated as shown:

The amount of heat released for formation of mole water = 1395 J / (25.0 m mol x 1mol/1000 m mol)

= 55,800 J