1. A metal sample weighing 124.10 g and at a temperature of 99.3°C was placed in
ID: 893689 • Letter: 1
Question
1. A metal sample weighing 124.10 g and at a temperature of 99.3°C was placed in 42.87 g of water in a calorimeter at 23.9°C. At equilibrium the temperature of the water and metal was 41.6°C.
What was t for the water? (t = tfinal tinitial)
What was t for the metal?
How much heat flowed into the water?
____________ °C
____________ °C
____________ joules
Taking the specific heat of water to be 4.18 J/g°C, calculate the specific heat of the metal, using Equation 3.
What is the approximate molar mass of the metal? (Use Eq. 4.)
____________ joules/g°C
____________ g
2. When 5.12 g of NaOH were dissolved in 51.55 g water in a calorimeter at 24.5°C, the temperature of the solution went up to 49.8°C.
Is this solution reaction exothermic? ____________ Why?
Calculate qH2O , using Equation 1.
Find H for the reaction as it occurred in the calorimeter (Eq. 5).
____________ joules
H = ____________ joules (continued on following page)
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104 Experiment 14 Heat Effects and Calorimetry
d. Find H for the solution of 1.00 g NaOH in water.
e. Find H for the solution of 1 mole NaOH in water.
H = ____________ joules/mole
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.
Given the following heats of formation, Hf , in kJ per mole, as obtained from a table of Hf data, calculate H 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
H = ____________ kJ
Explanation / Answer
1. A metal sample weighing 124.10 g and at a temperature of 99.3°C was placed in 42.87 g of water in a calorimeter at 23.9°C. At equilibrium the temperature of the water and metal was 41.6°C.
What was t for the water? (t = tfinal tinitial)
Solution :- t = tfinal – tinitial
= 41.6 C – 23.9 C
= 17.7 C
What was t for the metal?
Solution :- t = tfinal – tinitial
= 41.6 C – 99.3 C
= -57.7 C
How much heat flowed into the water?
Solution : -
Formula
q=m*c*delta T
= 42.87 g * 4.18 J per g C * 17.7 C
= 3172 J
Therefore heat flown in water = 3172 J
Taking the specific heat of water to be 4.18 J/g°C, calculate the specific heat of the metal, using Equation 3.
Solution :-
The amount of heat absorbed by water is same as the amount of heat lost by the metal so q =3172 J
q= m*c*delta T
c= q/ m*delta T
= 3172 J / 124.10 g * 57.7 C
= 0.443 J per g C
Therefore specific heat of metal is 0.443 J per g C
What is the approximate molar mass of the metal? (Use Eq. 4.)
Solution :- molar mass = 25 J /mol C/ specific heat
= 25 J per mol C / 0.443 J per g
= 56.43 g per mol
Therefore the molar mass of the metal is 56.43 g
2. When 5.12 g of NaOH were dissolved in 51.55 g water in a calorimeter at 24.5°C, the temperature of the solution went up to 49.8°C.
Is this solution reaction exothermic? ____________ Why?
Solution :- Reaction is exothermic because temperature is increasing means reaction gives off heat
Calculate qH2O , using Equation 1.
Solution :-
qwater = m*c*delta T
= 51.55 g * 4.18 J per g C * (49.8 C- 24.5C)
= 5452 J
Find H for the reaction as it occurred in the calorimeter (Eq. 5)
Solution : - Delta H reaction = -5452 J
Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
104 Experiment 14 Heat Effects and Calorimetry
d. Find H for the solution of 1.00 g NaOH in water.
Solution :- 5.12 g NaOH = 5452 J
1.0g NaOH * 5452 J / 5.12 g = -1065 J per g NaOH
e. Find H for the solution of 1 mole NaOH in water.
Solution :- 1 mol NaOH = 40 g
40 g NaOH * 1065 J / 1 g = -42600 J/mol
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.
Given the following heats of formation, Hf , in kJ per mole, as obtained from a table of Hf data, calculate H 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
Solution :-
Delta Hrxn = sum of delta H product – sum of delta H reactant
= [(Na^+ *1)+(OH^-*1)]-[NaOH*1]
= [(-240.1*1)+(-230.0*1)] –[-425.6*1]
= -44.5 kJ per mol
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