When a student mixes 50 ml of 1.0 m hcl and 50 ml of 1.0 m naoh in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 °c to 27.5 °c. calculate the enthalpy change for the reaction in kj per mol of hcl, assuming that the calorimeter loses only a negligible quantity of heat. the total volume of the solution is 100 ml, its density is 1.0 g/ml, and its specific heat is 4.18 j/g*k.

54.4 kJ/mol

Explanation:

First we have to calculate the moles of HCl and NaOH.

The balanced chemical reaction will be,

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.05 mole of HCl neutralizes by 0.05 mole of NaOH

Thus, the number of neutralized moles = 0.05 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 =

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

where,

q = heat absorbed = ?

= specific heat of water =

m = mass of water = 100 g

= final temperature of water =

= initial temperature of metal =

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

Thus, the heat released during the neutralization = 2.72 KJ

Now we have to calculate the enthalpy of neutralization per mole of :

0.05 moles of releases heat = 2.72 KJ

1 mole of releases heat =

Thus the enthalpy change for the reaction in kJ per mol of HCl is 54.4 kJ

the answer is 303.15 or 303