A quantity of 2.00 × 102 mL of 0.461 M HCl is mixed with 2.00 × 102 mL of 0.231 M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the HCl and Ba(OH)2 solutions is the same at 21.66°C. For the process below, the heat of neutralization is −56.2 kJ/mol. What is the final temperature of the mixed solutions? H+(aq) + OH−(aq) → H2O(l)

Respuesta :

Answer : The final temperature of the mixed solutions will be, [tex]24.76^oC[/tex]

Explanation :

First we have to calculate the moles of [tex]HCl[/tex] and [tex]Ba(OH)_2[/tex].

[tex]\text{Moles of }HCl=\text{Molarity of }HCl\times \text{Volume of solution}=0.461mole/L\times 0.2L=0.0922mole[/tex]

[tex]\text{Moles of }Ba(OH)_2=\text{Molarity of }Ba(OH)_2\times \text{Volume of solution}=0.231mole/L\times 0.2L=0.0462mole[/tex]

Now we have to calculate the limiting reactant.

The balanced chemical reaction will be,

[tex]2HCl+Ba(OH)_2\rightarrow BaCl_2+2H_2O[/tex]

As, 1 mole of [tex]Ba(OH)_2[/tex] react with 2 mole of HCl

So, 0.0462 mole of [tex]Ba(OH)_2[/tex] react with [tex]2\times 0.0462=0.0924[/tex] mole of HCl

From this we conclude that, [tex]HCl[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]Ba(OH)_2[/tex] is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of water.

From the balanced chemical reaction, we conclude that

As, 1 mole of [tex]Ba(OH)_2[/tex] react to give 2 moles of [tex]H_2O[/tex]

So, 0.0462 mole of [tex]Ba(OH)_2[/tex] react to give [tex]2\times 0.0462=0.0924[/tex] moles of [tex]H_2O[/tex]

The moles of water = 0.0924 mole

Now we have to calculate the heat released.

As, 1 mole of water releases heat = 56.2 KJ/mole

So, 0.0924 mole of water releases heat = [tex]0.0924\times 56.2KJ/mole=5.19288KJ=5192.88J[/tex]

Now we have to calculate the final temperature of solution.

[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]

where,

q = amount of heat = 5192.88 J

[tex]c[/tex] = specific heat capacity = [tex]4.18J/g^oC[/tex]

The volume of water = 200 ml + 200 ml = 400 ml

As the density of water is, 1 g/ml.

So, the mass of water = [tex]1g/ml\times 400ml=400g[/tex]

[tex]T_{final}[/tex] = final temperature = ?

[tex]T_{initial}[/tex] = initial temperature = [tex]21.66^oC[/tex]

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

[tex]5192.88J=400g\times 4.18J/g^oC\times (T_{final}-21.66)^oC[/tex]

[tex]T_{final}=24.76^oC[/tex]

Therefore, the final temperature of the mixed solutions will be, [tex]24.76^oC[/tex]