Question

HCl: -164.4 kJ/mol

NaOH: -469.6 kJ/mol

NaCl: -407.1 kJ/mol

H2O: -285.9 kJ/mol

If 50 mL of 1.00 M HCl and 50 mL of 2.00 M NaOH are mixed, given the enthalpies of formation above, use Hess's Law to calculate the molar heat of neutralization.

Answer 1

HCl + NaOH NaCl + H2O H =?

No of moles of HCl in one litre = Volume x Molarity / 1000

= 50 x 1.00 / 1000 = 0.05 Moles

No of moles in of NaOH in one litre = 50 x 2.00 / 1000

= 0.1 moles

H⁺ + Cl^- HCl H1 = - 164.4 kj/mol

Na⁺ + OH ^- NaOH  H2 = - 469.6 kj/ mol

Na⁺ + Cl^- NaCl  H3 = - 407.1 kj/mol

H⁺ + OH^- HCl H4 = - 285.9 kj/mol

All the enthalpy values given for combination ions to produce ionic compounds. In this process energy will be released so that net energy change is negative.

Net energy Change in the single step process is equal to the net energy change in the several step process.

H = H1 + H2 + H3 + H4

= 164.4 + 469.6 - 407.1 - 285.9

= 59 kj/mol  

Only 0.05 Moles of HCl with low concentration can be neutralized by 0.05 moles of NaOH. Remaining NaOH will be remained as such.

So that the energy change in one mole process = 59 kj

The energy change in 0.05 moles process how much =?

= 59 x 0.05

= 2.59 Kj