DATA ON BOTTOM OF PAGE Must Use This Data Mass of NaOH: 4.0294g Mass of H20: 100
ID: 504905 • Letter: D
Question
DATA ON BOTTOM OF PAGE
Must Use This Data
Mass of NaOH: 4.0294g
Mass of H20: 100g
pH: 13.0
Temperature in C Time in Seconds 20 0 21 10 21 20 22 30 24 40 25 50 25 60 25 70 26 80 27 90 28 100 28 110 DATA ANAL TSIS 1. Using the data you recorded on the previous page, plot 3 graphs of Temperature (°C) vs. Time (sec) for reactions 1, and 3. May use either Excel or Graphical Analysis program 2. Reaction 1: Dissolution of NaOH mass of solution: B Calculate Heat (q Enthalpy (AH) of moles (n) Enthalpy in kJ/mol:Explanation / Answer
We have;
m(NaOH) = mass NaOH(s) = 4.0294 g
m(H2O) = mass H2O(l) = 100 g
T = temperature increase = 8.0oC
cg(H2O) = specific heat capacity of water = 4.18 JoC-1g-1
m = mass of solvent + mass of solute in grams = 100 + 4.0294 = 104.0294 g
heat released = q = m x cg x T = 104.0294 g x 4.18 JoC-1g-1 x 8.0oC = 3478.74 J
moles of solute = n(NaOH) = mass(NaOH) ÷ M(NaOH)
m(NaOH) = 4.0294 g
M(NaOH) = 22.99 + 16.00 + 1.008 = 39.998 g mol-1
n(NaOH) = 4.0294 ÷ 39.998 =0.100 mol
Hsoln = q ÷ n(NaOH)
Hsoln is negative (process is exothermic)
q = 3478.74 J
n(NaOH) = 0.100 mol
Hsoln = -3478.74 ÷ 0.100 = -34787.4 J mol-1
Hsoln = - 34787.4 J ÷ 1000 J/kJ = -34.79 kJ/mol
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