SHOW ALL WORK! Place 50.0 mL each of 1.0 M HCl and 1.0 M NaOH at room temperatur

Question

SHOW ALL WORK!

Place 50.0 mL each of 1.0 M HCl and 1.0 M NaOH at room temperature in respective calorimeters for 3 mintues:

HCl- inital average temp before mixing 22.2*C

NaOH-inital average tem before mixting 23*C

4th min: HCl is mixed with NaOH

5th-20th min: 25.6, 25.8, 26, 26.3, 26.5, 26.8, 27.1, 27.4, 27.7, 28, 28.3, 28.5, 28.8, 29, 29.3, 29.6

(*for Temperature gained, use the temperature minus the average temperature of HCl+NaOH. The density of 0.5 M NaCL produced is 1.02 g/mL, and its specific heat is 4.00 Jg^-1C^-1)

What is the heat gaining by mixed solutions?

Heat of the reaction?(negative of heat gained)

Enthalpy of neutralization (per mole of water formed)

Explanation / Answer

Average temperature of mixed solution = (22.2+23)/2 =22.6 deg.c

Toral volume of solution= 50+50 =100ml

Density of NaCl =100*1.02 g/ml= 102 gm

This is the mass of NaCl formed=

Heat gained= Mass*Specific heat* Temperature differece= 102*(29.6( final temperature)- 22.6)*4=2856 joules

The reaction is HCl+ NaOH----> NaCl+ H2O

moles of HCl required= 1*50/1000=0.05 = moles of NaOH required= moles of NaCl formed= 0.05

So heat gained= 2856/0.05 J/mol =57120 j/mol

Heat of reaction = -57120 j/mol

since moles of water is same that of moles of NaCl formed,

Enthalpy of neutralization= -57120 j/mol

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