In a certain calorimeter, the reaciton between 100cm3 of 0.1M HCL and 100cm3 of 0.1M NaOH causes temp rise of?

Question

cont:
0.70 C. Assuming no heat losses, calculate: the heat of reaction for H(aq) + OH(aq) -> H2O.
(For the resulting solution - density = 1.0gcm-3 and specific heat = 4.18.)

Thanks heaps in advance!

0.70 C. Assuming no heat losses, calculate: the heat of reaction for H(aq) + OH(aq) -> H2O.
(For the resulting solution - density = 1.0gcm-3 and specific heat = 4.18.)

Ok here's what I have:
If I add 100cm3 of HCl and 100cm3 of NaOH final solution = 200cm3 which is:
Density = mass/volume
1.00g/cm3 = m x 200cm3
200 g = m

Therefore:
(delta) H = mc(delta)T
(delta) H= 200g x 4.18J/g/K x 0.70K
(delta) H= -585.2 J/mol

But the answer sheet says its -50kJ/mol
where have I gone wrong?

Hi Gervald... Thanks for your answer, but
how do you get 0.01 mole of reactant?
Thanks

Answer 1

The ΔH you calculated above is the heat required to raise the temperature of the resulting solution by .7 °C. The ΔH you found is very specific to this case and should be in units of Joules (or kJ). This heat is provided by a neutralization reaction between 100 mL of .1 M HCl and 100 mL of .1 M NaOH. If you take into account how many moles of reactants are involved in the specific case you solved for and adjust your answer accordingly, you can get a molar heat of the reaction.

Molarity = moles of solute / Liters of solution
Solving for moles we get,
Moles of solute = Molarity * Liters
Both solutions are .1 Molar and have a volume of .1 Liter.
Moles = .1 M * .1 L = .01 moles

So only .01 moles of H+ and OH- are reacting in this problem and (according to your calculations) 585.2 Joules are released. If you divide the Energy released by the number of moles reacted you will get the ΔH value per mole you desire.

ΔH = .585 kJ / .01 moles
ΔH = 58.5 kJ per mol

Answer 2

Your first answer should be in J, not J/mol.
You have 0.01 mol of each reactant, and therefore you divide 500 (or so) by 0.1 to get 50kJ/mol.
Then you put the negative sign in, to make it exothermic.