- Calculate the percentage error between the value calculated in question 3.a (theoretical value) and the value calculated in question 3.b (experimental value). [3.98 % ]
- Calculate the percentage error between the value calculated in question 3.a (theoretical value) and the value calculated in question 3.b (experimental value). [3.98 % ]
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter5: Thermochemistry
Section: Chapter Questions
Problem 5.99QE
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States of Matter
The substance that constitutes everything in the universe is known as matter. Matter comprises atoms which in turn are composed of electrons, protons, and neutrons. Different atoms combine together to give rise to molecules that act as a foundation for all kinds of substances. There are five states of matter based on their energies of attraction, namely solid, liquid, gases, plasma, and BEC (Bose-Einstein condensates).
Chemical Reactions and Equations
When a chemical species is transformed into another chemical species it is said to have undergone a chemical reaction. It consists of breaking existing bonds and forming new bonds by changing the position of electrons. These reactions are best explained using a chemical equation.
Question
![3. Use the information contained in "Lab Exercise 11.B" on page 506 in the textbook to
answer the following questions.
a. Use the reaction equations and reaction enthalpies shown in the introduction section to
calculate the following:
CSH12CL)-> 5C(s) -
6H2009) 6420 LL
A
• a reaction enthalpy for the combustion of pentane
• a molar enthalpy of combustion for pentane
[ 3488.7 kJ and 3488.7 kJ/mol ]
Answer:
CsHizW180₂(g)-56₂9³ +64₂0(1)
CSH12C1-35CCs) +6 H₂(g)
H₂Ocg)-4₂0x6
6H20c9-620(1) @4=90.65
*5cc)+502) -560₂ (9)
*6=64₂ (9)+302(g)-(20(g
0
Answer:
b. Use the calorimetric data provided to calculate a molar enthalpy of combustion for
pentane
56cs)+50₂(g) - 500₂6g
6H₂(g) +30₂9-6H₂0lg
Answer:
-3488.7hJ
I mol
[3.35 x10 kJ/mol ]
=-mclt
= -mcat/n
=C1.24 kg (4.197( 37.6-18.4)(0.0297907718mol)
-3.35×103 kJ/mol
[3.98 % ]
-=-3488.7 K
c. Calculate the percentage error between the value calculated in question 3.a
(theoretical value) and the value calculated in question 3.b (experimental value).
0
4. Use the information contained in "Lab Exercise 11.D" on page 513 in the textbook to
answer the following questions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb5fdd233-e947-4d5f-a3c0-4edca06f47b3%2F6117204b-d893-4193-bacc-4bf7f2302f32%2F8k2e7wp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:3. Use the information contained in "Lab Exercise 11.B" on page 506 in the textbook to
answer the following questions.
a. Use the reaction equations and reaction enthalpies shown in the introduction section to
calculate the following:
CSH12CL)-> 5C(s) -
6H2009) 6420 LL
A
• a reaction enthalpy for the combustion of pentane
• a molar enthalpy of combustion for pentane
[ 3488.7 kJ and 3488.7 kJ/mol ]
Answer:
CsHizW180₂(g)-56₂9³ +64₂0(1)
CSH12C1-35CCs) +6 H₂(g)
H₂Ocg)-4₂0x6
6H20c9-620(1) @4=90.65
*5cc)+502) -560₂ (9)
*6=64₂ (9)+302(g)-(20(g
0
Answer:
b. Use the calorimetric data provided to calculate a molar enthalpy of combustion for
pentane
56cs)+50₂(g) - 500₂6g
6H₂(g) +30₂9-6H₂0lg
Answer:
-3488.7hJ
I mol
[3.35 x10 kJ/mol ]
=-mclt
= -mcat/n
=C1.24 kg (4.197( 37.6-18.4)(0.0297907718mol)
-3.35×103 kJ/mol
[3.98 % ]
-=-3488.7 K
c. Calculate the percentage error between the value calculated in question 3.a
(theoretical value) and the value calculated in question 3.b (experimental value).
0
4. Use the information contained in "Lab Exercise 11.D" on page 513 in the textbook to
answer the following questions.
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