Question

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.)

Answer 1

Answer:

The ΔH of this reaction is 55.8 kJ/mol.

Explanation:

$Molarity=\frac{Moles}{Volume}$

Moles of sulfuric acid:

Moles of sulfuric acid = $0.500 mol/L\times 0.026 L=0.013 mol$

Moles of Potassium hydroxide:

Moles of Potassium hydroxide= $1.00 mol/L\times 0.026 L=0.026 mol$

$H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O,\Delta H_r=?$

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

Then 0.013 mol of  sulfuric acid will react with:

$0.013\times \frac{2}{1}=0.026 mol$ 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=$Density\times Volume=1.00 g/mL\times 52.0 ml=52.000 g$

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

$Q=mc\Delta T=52.000 g\times 4.184 J/g^oC\times 6.67^oC$

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:

$\frac{1.4511 kJ}{0.026 mol}=55.8 kJ$

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