Question

Suppose 5.0 mL of 3.0 HCl is added to 8.0 mL of 2.5M NaOH, change in temp 4.3

find change H rxn c=4.18 J/g

Please show work/steps for study purposes. Thank you in advance!

Answer 1

We know that,

Moles = Molarity x Volume (in L)

=> Moles of HCl added = 3 x 0.005 = 0.015

Moles of NaOH added = 2.5 x 0.008 = 0.02

The reaction is,

HCl + NaOH -----> NaCl + H2O

According to the stoichiometry of the reaction 1 mole of HCl reacts with 1 mole of NaOH

=> Moles of HCl reacted = 0.015 = Moles of NaOH reacted.

Moles of NaOH left over = 0.02 - 0.015 = 0.005 moles

Total volume of solution = 5 + 8 = 13 mL

Assumption: Density of Solution = 1 g / mL

=> Mass of solution = 13 g

Cp = 4.18

delta T = 4.3

Heat released by reaction = m Cp delta T

=> Heat = 13 x 4.18 x 4.3 = 233.662 J

=> Heat released when 0.015 moles of reactant react = 233.662 J

=> Heat released when 1 mole of reactants react = 233.662 / 0.015 = 15577.47 J

=> delta H = - 15577.47 J / mol = - 15.577 kJ / mol (-ve since heat is being released)