Question

In a constant‑pressure calorimeter, 70.0 mL of 0.770 M H2SO4 is added to 70.0 mL of 0.420 M NaOH. The reaction caused the temperature of the solution to rise from 23.14 ∘C to 26.00 ∘C. If the solution has the same density and specific heat as water ( 1.00 g/mL and 4.184 J/(g⋅°C), respectively), what is ΔH for this reaction (per mole of H2O produced)

Answer 1

Answer : The enthalpy of neutralization is, -113.9 KJ/mole

Explanation :

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

$\text{Moles of }H_2SO_4=\text{Concentration of }H_2SO_4\times \text{Volume of solution}=0.770mole/L\times 0.070L=0.0539mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.420mole/L\times 0.070L=0.0294mole$

The balanced chemical reaction will be,

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

From the balanced reaction we conclude that,

As, 2 mole of $NaOH$ neutralizes by 1 mole of $H_2SO_4$

As, 0.0294 mole of $NaOH$ neutralizes by $\frac{0.0294}{2}=0.0147$ mole of $H_2SO_4$

Thus, the number of neutralized moles = 0.0147 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 = $70.0ml+70.0ml=140.0ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 140.0ml=140.0g$

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 = 140.0 g

$T_{final}$ = final temperature of water = $26.00^oC$

$T_{initial}$ = initial temperature of metal = $23.14^oC$

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

$q=140.0g\times 4.184J/g^oC\times (26.00-23.14)^oC$

$q=1675.27J=1.675kJ$

Thus, the heat released during the neutralization = -1.675 KJ

Now we have to calculate the enthalpy of neutralization.

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

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -1.675 KJ

n = number of moles used in neutralization = 0.0147 mole

$\Delta H=\frac{-1.675KJ}{0.0147mole}=-113.9KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, -113.9 KJ/mole