Textbook Question
Consider the decomposition of liquid benzene, C6H6(l), to gaseous acetylene, C2H2(g): C6H6(l) → 3 C2H2(g) ΔH = +630 kJ (c) Which is more likely to be thermodynamically favored, the forward reaction or the reverse reaction?
Ch.5 - Thermochemistry
Brown15th EditionChemistry: The Central ScienceISBN: 9780137542970
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch.1 - Introduction: Matter, Energy, and Measurement146
- Ch.2 - Atoms, Molecules, and Ions200
- Ch.3 - Chemical Reactions and Reaction Stoichiometry176
- Ch.4 - Reactions in Aqueous Solution155
- Ch.5 - Thermochemistry126
- Ch.6 - Electronic Structure of Atoms131
- Ch.7 - Periodic Properties of the Elements126
- Ch.8 - Basic Concepts of Chemical Bonding123
- Ch.9 - Molecular Geometry and Bonding Theories143
- Ch.10 - Gases147
- Ch.11 - Liquids and Intermolecular Forces91
- Ch.12 - Solids and Modern Materials122
- Ch.13 - Properties of Solutions126
- Ch.14 - Chemical Kinetics174
- Ch.15 - Chemical Equilibrium101
- Ch.16 - Acid-Base Equilibria139
- Ch.17 - Additional Aspects of Aqueous Equilibria146
- Ch.18 - Chemistry of the Environment72
- Ch.19 - Chemical Thermodynamics129
- Ch.20 - Electrochemistry142
- Ch.21 - Nuclear Chemistry101
- Ch.22 - Chemistry of the Nonmetals68
- Ch.23 - Transition Metals and Coordination Chemistry66
- Ch.24 - The Chemistry of Life: Organic and Biological Chemistry10
Brown15th EditionChemistry: The Central ScienceISBN: 9780137542970Not the one you use?Change textbook
Chapter 5, Problem 47b
Consider the combustion of isopropanol, C3H7OH(l), which is the primary component of rubbing alcohol: C3H7OH(l) + 9/2 O2(g) → 3 CO2(g) + 4 H2O(l) ΔH = -2248 kJ (b) Balance the forward reaction with whole-number coefficients. What is ΔH for the reaction represented by this equation?
Verified step by step guidance1
Identify the given unbalanced chemical equation: \( \text{C}_3\text{H}_7\text{OH}(l) + \frac{9}{2} \text{O}_2(g) \rightarrow 3 \text{CO}_2(g) + 4 \text{H}_2\text{O}(l) \).
To balance the equation with whole-number coefficients, multiply all coefficients by 2 to eliminate the fraction: \( 2 \text{C}_3\text{H}_7\text{OH}(l) + 9 \text{O}_2(g) \rightarrow 6 \text{CO}_2(g) + 8 \text{H}_2\text{O}(l) \).
Verify that the number of atoms of each element is the same on both sides of the equation: 6 C, 16 H, and 18 O on both sides.
Since the coefficients of the balanced equation are doubled, the enthalpy change \( \Delta H \) for the reaction must also be doubled. Calculate the new \( \Delta H \) by multiplying the given \( \Delta H = -2248 \text{ kJ} \) by 2.
The new \( \Delta H \) represents the enthalpy change for the balanced reaction with whole-number coefficients.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
1mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Combustion Reactions
Combustion reactions involve the reaction of a substance with oxygen, producing heat and light. In organic compounds, such as isopropanol, combustion typically yields carbon dioxide and water. Understanding the stoichiometry of combustion is essential for balancing the reaction and calculating energy changes.
Recommended video:
Guided course
02:24
Combustion Apparatus
Balancing Chemical Equations
Balancing chemical equations ensures that the number of atoms for each element is the same on both sides of the equation. This is crucial for accurately representing the conservation of mass during a chemical reaction. Whole-number coefficients are used to achieve this balance, which is necessary for stoichiometric calculations.
Recommended video:
Guided course
01:32Balancing Chemical Equations
Enthalpy Change (ΔH)
Enthalpy change (ΔH) represents the heat absorbed or released during a chemical reaction at constant pressure. A negative ΔH indicates an exothermic reaction, where heat is released, as seen in combustion. Understanding ΔH is vital for evaluating the energy efficiency of reactions and predicting the feasibility of chemical processes.
Recommended video:
Guided course
02:34Enthalpy of Formation
Related Practice
Textbook Question
At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO3: 2 KClO3(s) → 2 KCl(s) + 3 O2(g) ΔH = -89.4 kJ For this reaction, calculate H for the formation of (b) 10.4 g of KCl.
Textbook Question
Consider the decomposition of liquid benzene, C6H6(l), to gaseous acetylene, C2H2(g): C6H6(l) → 3 C2H2(g) ΔH = +630 kJ (a) What is the enthalpy change for the reverse reaction?
Textbook Question
Consider the decomposition of liquid benzene, C6H6(l), to gaseous acetylene, C2H2(g): C6H6(l) → 3 C2H2(g) ΔH = +630 kJ (b) What is H for the formation of 1 mol of acetylene?
Textbook Question
At one time, a common means of forming small quantities of oxygen gas in the laboratory was to heat KClO3: 2 KClO3(s) → 2 KCl(s) + 3 O2(g) ΔH = -89.4 kJ c. Now consider the reverse reaction, in which KClO3 is formed from KCl and O2. What is Δ𝐻 for the formation of 19.1 g KClO3 from KCl and O2?
Textbook Question
Consider the combustion of isopropanol, C3H7OH(l), which is the primary component of rubbing alcohol: C3H7OH(l) + 9/2 O2(g) → 3 CO2(g) + 4 H2O(l) ΔH = -2248 kJ a. What is the enthalpy change for the reverse reaction?