Question

When 22.7 mL of 0.500 M H2SO4 is added to 22.7 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.)___kJ/mol H2O

Answer 1

Answer : The enthalpy of neutralization is, -55.8 KJ/mole

Explanation :

First we have to calculate the moles of $H_2SO_4$ and KOH.

$\text{Moles of }H_2SO_4=\text{Concentration of }H_2SO_4\times \text{Volume of solution}=0.500mole/L\times 0.0227L=0.01135mole$

$\text{Moles of KOH}=\text{Concentration of KOH}\times \text{Volume of solution}=1.00mole/L\times 0.0227L=0.0227mole$

Now we have to calculate the limiting reagent.

The balanced chemical reaction will be,

$H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O$

From the balanced reaction we conclude that,

As, 1 mole of $H_2SO_4$ react to give 2 moles of $H_2O$

So, 0.01135 mole of $H_2SO_4$ react to give $2\times 0.01135=0.0227$ moles of $H_2O$

and,

As, 2 moles of $KOH$ react to give 2 moles of $H_2O$

As, 0.0227 moles of $KOH$ react to give 0.0227 moles of $H_2O$

From this we conclude that they can form the same amount of product. This means that both will consume completely in the reaction.

Thus, the number of neutralized moles = 0.0227 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $22.7ml+22.7ml=45.4ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 45.4ml=45.4g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.184J/g^oC$

m = mass of water = 45.4 g

$T_{final}$ = final temperature of water = $30.17^oC$

$T_{initial}$ = initial temperature of water = $23.50^oC$

Now put all the given values in the above formula, we get:

$q=45.4g\times 4.184J/g^oC\times (30.17-23.50)^oC$

$q=1266.99J=1.267kJ$

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -1.267 KJ

n = number of moles used in neutralization = 0.0227 mole

$\Delta H=\frac{-1.267KJ}{0.0227mole}=-55.8KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, -55.8 KJ/mole