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* \text{Volume of solution}=0.770mole/L* 0.070L=0.0539mole`

`\text{Moles of NaOH}=\text{Concentration of NaOH}* \text{Volume of solution}=0.420mole/L* 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
`(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}* \text{Volume of water}=1g/ml* 140.0ml=140.0g`

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

`q=m* c* (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* 4.184J/g^oC* (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=(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=(-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