Calorimetry Example: Understanding AHn rxn If 100.0 mL 1.00 M HCI and 100.0 mL 1.00 M NAOH mixed: HCI(aq) + NAOH(aq) H20(1) + NaCI(aq) - Csoln = Cwater = 4.184 J/(g °C) - dsoln = dwater = 1.00 g/mL - mgoln = (100 mL + 100 mL) = 200 mL = 200 g AT soln = 6.9 °C (twice the water, twice the moles) 9soln = (200 g)(4.184 J/(g °C)(6.9 °C) = +5773.92 J FMoles HCI: (M*L) = (1.00 M)(0.100 L) = 0.100 moles Moles NaOH = (1.00 M)(0.100 L) = 0.100 moles 9soln = +5773.92 J / 0.100 moles = +57,739.2 J J/mole %3D %3D %3D %3D | %3D %3D - %3D %3D %3D %3D %3D %3D AHn = -57.7 kJ/mole rxn
Calorimetry Example: Understanding AHn rxn If 100.0 mL 1.00 M HCI and 100.0 mL 1.00 M NAOH mixed: HCI(aq) + NAOH(aq) H20(1) + NaCI(aq) - Csoln = Cwater = 4.184 J/(g °C) - dsoln = dwater = 1.00 g/mL - mgoln = (100 mL + 100 mL) = 200 mL = 200 g AT soln = 6.9 °C (twice the water, twice the moles) 9soln = (200 g)(4.184 J/(g °C)(6.9 °C) = +5773.92 J FMoles HCI: (M*L) = (1.00 M)(0.100 L) = 0.100 moles Moles NaOH = (1.00 M)(0.100 L) = 0.100 moles 9soln = +5773.92 J / 0.100 moles = +57,739.2 J J/mole %3D %3D %3D %3D | %3D %3D - %3D %3D %3D %3D %3D %3D AHn = -57.7 kJ/mole rxn
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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For calculating delta h, in the example it shows what to do if the reactants have a one to one ratio. How do we normalize if it isn't a one to one ratio?

Transcribed Image Text:Calorimetry Example: Understanding AHn
rxn
If 100.0 mL 1.00 M HCI and 100.0 mL 1.00 M NaOH mixed:
HCI(aq) + NaOH(aq) →
H20(1) + NaCl(aq)
- Csoln = Cwater = 4.184 J/(g °C)
-dsoln = dwater = 1.00 g/mL
msoln = (100 mL + 100 mL) = 200 mL = 200 g
- AT soln = 6.9 °C (twice the water, twice the moles)
9soln = (200 g)(4.184 J/(g °C)(6.9 °C) = +5773.92 J
Moles HCl: (M*L) = (1.00 M)(0.100 L) = 0.100 moles
Moles NaOH = (1.00 M)(0.100 L) = 0.100 moles
%3D
%3D
%3D
%3D
ΔΤ.
%3D
%3D
%3D
%3D
%3D
9soln = +5773.92 J /0.100 moles = +57,739.2 J J/mole
%3D
AHn = -57.7 kJ/mole
%3D
rxn
O 2014 Pearson Education, Inc.
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