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You place 33.8 ml of 0.154 M Ba(OH)2 in a coffee-cup calorimeter at 25.00°C and add 72.5 ml of 0.648 M HCl, also at 25.00°C. After stirring, the final temperature is 29.22°X. {assume that the total volume is the sum of the individual volumes and that the final solution has the same density (1.00 g/ml) and specific heat capacity (4.184 J/g°C) as water}. Calculate the change in enthalpy, \DeltaΔH, of the reaction (in kJ/mol) of water formed. Enter the appropriate sign (+/-). Enter to 1 decimal place.

Answer :

Answer : The enthalpy of reaction [tex](\Delta H_{rxn})[/tex] is, -180.5 kJ/mole

Explanation :

First we have to calculate the mass of solution.

[tex]Mass=Density\times Volume[/tex]

Volume of solution = Volume of HCl + Volume of [tex]Ba(OH)_2[/tex]

Volume of solution = 33.8 mL + 72.5 mL

Volume of solution = 106.3 mL

Density of solution = 1 g/mL

[tex]Mass=1g/mL\times 106.3mL=106.3g[/tex]

The mass of solution is, 106.3 grams.

Now we have to calculate the heat released in the system.

Formula used :

[tex]Q=m\times c\times \Delta T[/tex]

or,

[tex]Q=m\times c\times (T_2-T_1)[/tex]

where,

Q = heat released = ?

m = mass = 106.3 g

[tex]C_p[/tex] = specific heat capacity of water = [tex]4.184J/g^oC[/tex]

[tex]T_1[/tex] = initial temperature  = [tex]25^oC[/tex]

[tex]T_2[/tex] = final temperature  = [tex]29.22^oC[/tex]

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

[tex]Q=106.3g\times 4.184J/g^oC\times (29.22-25)^ioC[/tex]

[tex]Q=1876.88J=1.877kJ[/tex]        (1 kJ = 1000 J)

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

[tex]\text{Moles of }Ba(OH)_2=\text{Concentration of }Ba(OH)_2\times \text{Volume of solution}[/tex]

[tex]\text{Moles of }Ba(OH)_2=0.154M\times 0.0338L=5.20\times 10^{-3}mol[/tex]

and,

[tex]\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of solution}[/tex]

[tex]\text{Moles of }HCl=0.648M\times 0.0725L=4.69\times 10^{-2}mol[/tex]

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

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

From the balanced reaction we conclude that

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

So, [tex]5.20\times 10^{-3}[/tex]  moles of [tex]Ba(OH)_2[/tex] react with [tex]5.20\times 10^{-3}\times 2=0.0104[/tex] moles of [tex]HCl[/tex]

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 [tex]H_2O[/tex]

From the reaction, we conclude that

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

So, [tex]5.20\times 10^{-3}[/tex]  moles of [tex]Ba(OH)_2[/tex] react with [tex]5.20\times 10^{-3}\times 2=0.0104[/tex] moles of [tex]H_2O[/tex]

Now we have to calculate the change in enthalpy of the reaction.

[tex]\Delta H_{rxn}=-\frac{q}{n}[/tex]

where,

[tex]\Delta H_{rxn}[/tex] = enthalpy of reaction = ?

q = heat of reaction = 1.877 kJ

n = moles of reaction = 0.0104 mole

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

[tex]\Delta H_{rxn}=-\frac{1.877kJ}{0.0104mole}=-180.5kJ/mole[/tex]

The negative sign indicates that the heat is released.

Therefore, the enthalpy of reaction [tex](\Delta H_{rxn})[/tex] is, -180.5 kJ/mole

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