Question

Please help with 1, 3, and 4 :)

A. ΔH_solution of NaOH_(s) Phase Change:
Mass of NaOH: 1.81 g
ΔT for reaction A : 6 °C

B. ΔH_solution for neutralization of HCl_(aq) and NaOH_(aq):
Volume of HCl: .05 L
Volume of NaOH: .05 L
Volume total: .10 L
Molarity of HCl: 2.08 M
Molarity of NaOH: 2.08 M
ΔT for reaction B: 12°C

C. ΔH for Reaction of HCl_(aq) and NaOH_(s):
Volume of HCl: .055 L
Volume of water: .045 L
Volume total: .1 L
Molarity of HCl: 2.08 M
Mass of NaOH: 4.3 g
ΔT for reaction C: 22°C

D. ΔH for Solution of KCl_(s) in water:
ΔT for reaction D: -3.5°C
Mass of KCl: 5 g
Volume of water: .05 L

B. AH for neutralization of HClaq and NaOHaq: Calculate the molarity of the resulting sodium chloride solution. 2.08 Submit Answer Incorrect. Tries 1/3 Previous Tries Enter AT from the graph. 12.0 °C You are correct. Your receipt no. is 157-9589 Loc89 Previous Tries Calculate the value (calories) for the heat of reaction. cal Submit Answer Incorrect. Tries 1/3 Previous Tries Calculate AH for neutralization. kJ mol-1 Tries 0/3 Submit Answer The overall reaction for this neutralization is: Incorrect Endothermic Correct: Exothermic You are correct. Previous Tries Your receipt no. is 157-6865

Answer 1

SOLN B

NaOH: moles = molarity * volume

2.08 M x (0.05 L) = 0.104 moles NaOH

same is the mole of HCl

hence both will neutralize each other

hence moles of NaCl forme = 0.104 moles

Molarity = moles / total volume = 0.104 / 0.1 = 1.04 M

Heat = mass x T change x specific heat

mass = moles * molar mass = 0.104 *40 = 4.16 g

Heat = 4.16 g x 12 C x 1.00 cal/gC = 49.92 cal

dH = heat / mol = 49.92 cal / 0.104 mol = 480 cal/mol (of HCl or NaOH)

1cal = 4.184 J

hence 480 cal = 2007 J or 2.007 kJ/mol