Question

In a coffee‑cup calorimeter, 65.0 mL of 0.890 M H 2 SO 4 was added to 65.0 mL of 0.260 M NaOH . The reaction caused the temperature of the solution to rise from 23.78 ∘ C to 25.55 ∘ C. If the solution has the same density as water (1.00 g/mL) and specific heat as water (4.184 J/g‑K), what is Δ H for this reaction (per mole of H 2 O produced)? Assume that the total volume is the sum of the individual volumes.

Answer 1

Answer : The enthalpy of neutralization is, 56.96 kJ/mole

Explanation :

First we have to calculate the moles of H₂SO₄ and NaOH.

$\text{Moles of }H_2SO_4=\text{Concentration of }H_2SO_4\times \text{Volume of solution}=0.890mole/L\times 0.065L=0.0578mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=0.260mole/L\times 0.065L=0.0169mole$

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 H₂SO₄

So, 0.0169 mole of NaOH neutralizes by 0.00845 mole of H₂SO₄

That means, NaOH is a limiting reagent and H₂SO₄ is an excess reagent.

Now we have to calculate the moles of H₂O.

As, 2 mole of NaOH react to give 2 mole of H₂O

So, 0.0169 mole of NaOH react to give 0.0169 mole of H₂O

Now we have to calculate the mass of water.

As we know that the density of water is 1.00 g/ml. So, the mass of water will be:

The volume of water = $65.0ml+65.0ml=130ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1.00g/ml\times 130ml=130g$

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 = 130 g

$T_{final}$ = final temperature of water = $23.78^oC$

$T_{initial}$ = initial temperature of metal = $25.55^oC$

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

$q=130g\times 4.184J/g^oC\times (25.55-23.78)^oC$

$q=962.7J$

Thus, the heat released during the neutralization = -962.7 J

Now we have to calculate the enthalpy of neutralization.

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

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -962.7 J

n = number of moles used in neutralization = 0.0169 mole

$\Delta H=\frac{-962.7J}{0.0169mole}=-56964.49J/mole=-56.96kJ/mol$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 56.96 kJ/mole