Calculate the enthalpy of neutralization in kj/mol for HCl. ?

Question

This is the information they give:

the heat capacity of a certain calorimeter is 488.1J/ degrees C when containing 50 ml of water. When 25 mL of 0.700 M NaOH was mixed in the same calorimeter with 25 mL of 0.700 M HCl both initially at 20 degrees C, the temperature increased to 22.1 degrees C.

This is what I have so far.

qCal = 488.1 J/degrees C x 2.1 (change in temperature) = 1025.01 J

qWater = 50 grams H2O x 4.184 heat capacity x 2.1 = 439.32 J

the correct answer is 58.6 kJ/mol, but I dont know how to get there.

Answer 1

looking good, let's combine those energies:

1025.01 J +  439.32 J = 1454.33 Joules released

but by how many moles neutralized:

0.025 litres @ 0.700 mol/litre = 0.0175 moles nuetralized either acid or base making the same amount of water

so how much per 1 mole:

1454.33 joules / 0.0175 moles = 83,105 Joules

which is the wrong answer.... it's too big....

so let's modify...

===========================

what if the "calorimeter is 488.1J/ degrees C" included the 50 ml water,

then the total heat would only have been=

qCal = 488.1 J/degrees C x 2.1 (change in temperature) = 1025.01 J

then the per mole would be:

1025.01 J / 0.0175 moles = 58,572 Joules released

that's it

your answer is : dH neutralization is - 58.6 kJ