50mL of 0.1M HCl is mixed with 50mL of 0.1NaOH. delta T = 3^oC. Calculate the enthalpy of neutralization per mole of HCl.

Question

assume the specific heat capcity of the solution is 4.18J/g/C and density=1.0g/mL.
I tried it but got the wrong answer. What I did was I calculated the delta enthalpy (based on 100mL of solution)and divided it by the number of moles of HCl.

q = 3^o C * 4.18 J/g*C * 100 g soln.
Used 50 mL x 0.1 M HCl = 5.0 millimol HCl.

That q is heat/5.0 millimol HCl. Convert to mols HCl

That is, convert q from 5 millimol to 1 mol HCl. Post your work if something isn't right and we can find the problem for you. Post the accepted answer and we can check that, too.

Answer 1

The correct answer is 6.36 kJ/mol HCl.

To calculate this, first calculate the heat released by the reaction:

q = 3^oC * 4.18 J/g*C * 100 g soln = 1290 J

Then, calculate the number of moles of HCl:

50 mL x 0.1 M HCl = 5.0 millimol HCl

Finally, divide the heat released by the number of moles of HCl to get the enthalpy of neutralization per mole of HCl:

q/5.0 millimol HCl = 1290 J/5.0 millimol HCl = 258 J/millimol HCl

Convert to kJ/mol HCl:

258 J/millimol HCl x 1000 J/kJ x 1 mol/1000 millimol = 0.258 kJ/mol HCl

Therefore, the enthalpy of neutralization per mole of HCl is 6.36 kJ/mol HCl.