Question

Hess's Law:
Calculate ΔH for the neutralization of HCl_(aq) by NaOH_(s). I'M NOT SURE WHAT DATA YOU NEED, SO I COPY AND PASTED ALL OF MY DATA.

A. ΔH_solution of NaOH_(s) Phase Change:
Mass of NaOH: 2.05 g
ΔT for reaction A : 8.0 °C

B. ΔH_solution for neutralization of HCl_(aq) and NaOH_(aq):
Volume of HCl: 0.0500 L
Volume of NaOH: 0.0500 L
Volume total: 0.100 L
Molarity of HCl: 1.95 M
Molarity of NaOH: 1.95 M
ΔT for reaction B: 11.0°C

C. ΔH for Reaction of HCl_(aq) and NaOH_(s):
Volume of HCl: 0.0550 L
Volume of water: 0.0450 L
Volume total: 0.100 L
Molarity of HCl: 1.95 M
Mass of NaOH: 3.44 g
ΔT for reaction C: 17°C

D. ΔH for Solution of KCl_(s) in water:
ΔT for reaction D: -5.0°C
Mass of KCl: 5.02 g
Volume of water: 0.0500 L

Answer 1

B. ΔH_solution for neutralization of HCl_(aq) and NaOH_(aq):
Volume of HCl: 0.0500 L
Volume of NaOH: 0.0500 L
Volume total: 0.100 L
Molarity of HCl: 1.95 M
Molarity of NaOH: 1.95 M
ΔT for reaction B: 11.0°C

Solution :- Total volume = 0.100 L = 100 ml

Assuming the density of solution is 1 g /ml

Then mass of solution = 100 g

Now lets calculate the heat of reaction

q= m*s*delta T

s=specific heat of water = 4.184 J per g C

q=100 g * 4.184 J per g C * 11.0 C

q= 4602 J

now lets calculate the enthalpy change per mole

moles of NaOH = 1.95 mol per L * 0.05 L = 0.0975 mol

enthalpy change per mole NaOH = 4602 J / 0.0975 mol

                                                            = 47200 J per mol

47200 J * 1 kJ / 1000 J = 47.2 kJ per mol

It is written as -47.2 kJ per mol because exothermic reaction.

C. ΔH for Reaction of HCl_(aq) and NaOH_(s):
Volume of HCl: 0.0550 L
Volume of water: 0.0450 L
Volume total: 0.100 L
Molarity of HCl: 1.95 M
Mass of NaOH: 3.44 g
ΔT for reaction C: 17°C
moles of NaOH = 3.44 g / 40.0 g per mol = 0.086 mol

Moles of HCl = 1.95 mol per L * 0.055 L = 0.1073 mol HCl

Moles of NaOH are limiting

q= 100 g * 4.184 J per g C * 17.0 C

q= 7113 J

per mol NaOH

7113 J / 0.086 mol NaOH = 82709 J per mol

82709 J per mol * 1 kJ / 1000 J = 82.7 kJ per mol

So it is -82.7 kJ per mol Sign is negative because it is exothermic reaction

D. ΔH for Solution of KCl_(s) in water:
ΔT for reaction D: -5.0°C
Mass of KCl: 5.02 g
Volume of water: 0.0500 L

Solution :-

Moles of KCl = 5.02 g / 74.5513 g per mol = 0.06734 mol

Now lets calculate the q

q=m*s*delta T

q =50 g * 4.184 J per g C * 5.0 C

q= 1046 J

now lets convert it to per mol

1046 J / 0.06734 mol = 15533 J per mol

15533 J per mol * 1 kJ / 1000 J = 15.5 kJ per mol

So the delta H of solution of KCl is 15.5 kJ per mol