Question

You place 36.5 ml of 0.266 M Ba(OH)2 in a coffee-cup calorimeter at 25.00°C and add 56.6 ml of 0.648 M HCl, also at 25.00°C. After stirring, the final temperature is 29.83°C. {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, ΔH, of the reaction (in kJ/mol) of water formed. Enter the appropriate sign (+/-).

Answer 1

Answer : The enthalpy of reaction
`(\Delta H_(rxn))` is, -96.9 kJ/mole

Explanation :

First we have to calculate the mass of solution.

`Mass=Density* Volume`

Volume of solution = Volume of HCl + Volume of
`Ba(OH)_2`

Volume of solution = 56.6 mL + 36.5 mL

Volume of solution = 93.1 mL

Density of solution = 1 g/mL

`Mass=1g/mL* 93.1mL=93.1g`

The mass of solution is, 93.1 grams.

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

Formula used :

`Q=m* c* \Delta T`

or,

`Q=m* c* (T_2-T_1)`

where,

Q = heat released = ?

m = mass = 93.1 g

`C_p` = specific heat capacity of water =
`4.184J/g^oC`

`T_1` = initial temperature =
`25.0^oC`

`T_2` = final temperature =
`29.83^oC`

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

`Q=93.1g* 4.184J/g^oC* (29.83-25.00)^ioC`

`Q=1881.43J=1.88kJ` (1 kJ = 1000 J)

Now we have to calculate the moles of
`Ba(OH)_2` and HCl.

`\text{Moles of }Ba(OH)_2=\text{Concentration of }Ba(OH)_2* \text{Volume of solution}`

`\text{Moles of }Ba(OH)_2=0.266M* 0.0365L=9.71* 10^(-3)mol`

and,

`\text{Moles of }HCl=\text{Concentration of }HCl* \text{Volume of solution}`

`\text{Moles of }HCl=0.648M* 0.0566L=3.66* 10^(-2)mol`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O`

From the balanced reaction we conclude that

As, 1 mole of
`Ba(OH)_2` react with 2 mole of
`HCl`

So,
`9.71* 10^(-3)` moles of
`Ba(OH)_2` react with
`9.71* 10^(-3)* 2=0.0194` moles of
`HCl`

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

Now we have to calculate the moles of
`H_2O`

From the reaction, we conclude that

As, 1 mole of
`Ba(OH)_2` react to give 2 mole of
`H_2O`

So,
`9.71* 10^(-3)` moles of
`Ba(OH)_2` react with
`9.71* 10^(-3)* 2=0.0194` moles of
`H_2O`

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

`\Delta H_(rxn)=-(q)/(n)`

where,

`\Delta H_(rxn)` = enthalpy of reaction = ?

q = heat of reaction = 1.88 kJ

n = moles of reaction = 0.0194 mole

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

`\Delta H_(rxn)=-(1.88kJ)/(0.0194mole)=-96.9kJ/mole`

The negative sign indicates that the heat is released.

Therefore, the enthalpy of reaction
`(\Delta H_(rxn))` is, -96.9 kJ/mole