d. Find AH for the dissolution of 1.00 g NaOH in water. e. Find AH for the dissolution of 1 mole NaOH in water. ΔΗ = ΔΗ = kJ/g kJ/mol f. Given that NaOH exists as Na and OH ions in solution, write the equation for the reaction that occurs when NaOH is dissolved in water.
d. Find AH for the dissolution of 1.00 g NaOH in water. e. Find AH for the dissolution of 1 mole NaOH in water. ΔΗ = ΔΗ = kJ/g kJ/mol f. Given that NaOH exists as Na and OH ions in solution, write the equation for the reaction that occurs when NaOH is dissolved in water.
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter5: Principles Of Chemical Reactivity: Energy And Chemical Reactions
Section: Chapter Questions
Problem 113SCQ
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Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
Question
Can I get help solving question 2 sections d-f. Using the information from the fist page.
Transcribed Image Text:Experiment 14
Advance Study Assignment: Heat Effects and Calorimetry
1. A metal sample weighing 147.90 g and at a temperature of 99.5°C was placed in 49.73 g of water in a calo-
rimeter at 23.0°C. At equilibrium the temperature of the water and metal was 41.8°C.
What was Ar for the water? (Art final initial)
41.8°C -230°C
a.
b. What was Ar for the metal?
41.8-99.5
C.
How much heat flowed into the water? (Take the specific heat of the water to be 4.18 J/gºC.)
9 = SH. MAT > (4₁183 184) (49.73.8) (18.88) = 3908
-853383
e. What is the approximate molar mass of the metal? (Use Eq. 4.)
25
2007008
- 11/15°C) 458 45 ) =
MM =
d. Calculate the specific heat of the metal, using Equation 3.
9H₂0 = SH₁₂
•M. At
#20
=-SH₁ Mm Atm
3908 (SH)(147.⁹90) (-5.7.7) = +0.457=+5H (458
C.
b. Calculate 41120, using Equation 1.
90-418-3/5°c) (49.729/26.4%)=5487
Find AH for the reaction as it occurred in the calorimeter (Eq. 5).
AH = -9₂0
9=SH.M.At
M=4989+ 49.72 = 54.79
18,8 °C
-57.7c
2. When 4.98 g of NaOH was dissolved in 49.72 g of water in a calorimeter at 23.7°C, the temperature of the
solution went up to 50.1°C.
a.
Is this dissolution reaction exothermic? Yes
Why?
Because releases neat as its dissolved in water.
This raises the temperature of the Soution.
5.49x3 joules
9=154.79) (418-19) (26.400)
9=6036.3 J
-6036
AH =
391 x10³
joules
joules/g°C
54.6 mot
g/mol
6,04x103
joules
(continued on following page)
Transcribed Image Text:106 Experiment 14 Heat Effects and Calorimetry
d. Find AH for the dissolution of 1.00 g NaOH in water.
e.
Find AH for the dissolution of 1 mole NaOH in water.
ΔΗ Ξ
ΔΗ =
kJ/g
kJ/mol
f. Given that NaOH exists as Na and OH ions in solution, write the equation for the reaction that
occurs when NaOH is dissolved in water.
AH =
g. Given the following heats of formation, AH, in kJ per mole, as obtained from a table of AH, data,
calculate AH for the reaction in Part (f). Compare your answer with the result you obtained in Part (e).
NaOH(s), -425.6; Na (aq), -240.1; OH (aq), -230.0
kJ/mol
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