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 : The enthalpy of reaction is, -180.5 kJ/mole

Explanation :

First we have to calculate the mass of solution.

Volume of solution = Volume of HCl + Volume of

Volume of solution = 33.8 mL + 72.5 mL

Volume of solution = 106.3 mL

Density of solution = 1 g/mL

The mass of solution is, 106.3 grams.

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

Formula used :

or,

where,

Q = heat released = ?

m = mass = 106.3 g

= specific heat capacity of water =

= initial temperature  =

= final temperature  =

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

       (1 kJ = 1000 J)

Now we have to calculate the moles of and HCl.

and,

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 1 mole of react with 2 mole of

So,  moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of

From the reaction, we conclude that

As, 1 mole of react to give 2 mole of

So,  moles of react with moles of

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

where,

= 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:

The negative sign indicates that the heat is released.

Therefore, the enthalpy of reaction is, -180.5 kJ/mole