Consider the combustion of isopropanol, C 3 H 7 OH(l), which is the primary component of rubbing alcohol: C 3 H 7 OH(l) + 9/2 O 2 (g) → 3 CO 2 (g) + 4 H 2 O(l) ΔH = -2248 kJ (b) Balance the forward reaction with whole-number coefficients. What is ΔH for the reaction represented by this equation?

Hi everyone here we have a question asking us to write the balance forward reaction for the combustion of ice. So octane and calculate the entropy of the reaction. So we have S. O. Octane liquid plus 25 hats, oxygen gasses forms eight carbon dioxide gasses plus nine water liquid and our entropy is negative 5461 kg per mole. So we want whole numbers. So we're going to multiply this whole thing by two. So we're going to have to I so octane liquid plus oxygen gashes forms 16 carbon dioxide gashes plus water liquid. And because we multiplied it by two, we're going to multiply the entropy by 22. So our MPP is going to equal negative 10,000 kila jules Permal. And that is our final answer. Thank you for watching. Bye.

Ch.5 - Thermochemistry

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Brown 15th Edition

Ch.5 - Thermochemistry

Problem 47b

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?

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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.

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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.

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.

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.

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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

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?

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

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?

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?

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Key Concepts

Combustion Reactions

Balancing Chemical Equations

Enthalpy Change (ΔH)