A dilute solution of hydrochloric acid with a mass of 610 29 g and containing 0 33183 mol of HCl - Chemistry - Thermodynamics

Question

A dilute solution of hydrochloric acid with a mass of 610 29 g and
containing 0 33183 mol of HCl was exactly neutralized in a
calorimeter by the sodium hydroxide in 615 31 g of a dilute NaOH
solution The temperature increased from 16 784 to 20 610 °C The
specific heat of the HCl solution was 4 031 J g-1 °C-1; that of
the NaOH solution was 4 046 J g-1 °C-1 The heat capacity of the
calorimeter was 77 99 J °C-1 Use these data to calculate the
heat evolved by the following reaction
HCl(aq) + NaOH(aq) --> NaCl(aq) + H2O(l)

What is the heat of neutralization per mole of HCl? Assume that the
original solutions made independent contributions to the total heat
capacity of the system following their mixing

Lalit Kumar Sethia · Accepted Answer

Dear student, So I know: Î”T=3 829 degrees C Ccal=77
99J/degrees C CHCl=4 031J/g degrees C CNaOH=4 045J/g degrees C I'm
not positive in the next step I assumed that since the solutions
independently contributed to the total heat capacity that I do a
calculation for each so: Î”H surroundings=(77 99C/J x 3
826C) + (4 031J/gC x 610 29g x 3 826C) + (4 046J/gC x 615 31g x 3
826C)= 294 4J + 9412J + 9525J = 19235 4J so Î”H
reaction= -19234 4J To find the J/mol H+ I did -19235 4J/ 33183 mol
H+ since I was given 33183 mol HCl which = -57968J/mol H+ Regards