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If an equation for a reaction is doubled and then reversed, how is the value of \(\Delta H\) changed?

Short Answer

Expert verified
\( \Delta H \) becomes \(-2x\) after doubling and reversing the reaction.

Step by step solution

01

Understand the Initial Reaction and \( \Delta H \)

When we have a chemical reaction, the change in enthalpy (\( \Delta H \)) is provided or can be calculated for the reaction as written. The sign and magnitude of \( \Delta H \) depend on the particular reaction and the direction in which it is written.
02

Effect of Doubling the Reaction

Doubling a reaction means multiplying all the reactants and products by 2. When the equation of a reaction is doubled, the value of \( \Delta H \) is also doubled. For example, if the original \( \Delta H \) is \( x \), after doubling, \( \Delta H \) becomes \( 2x \).
03

Reverse the Reaction

Reversing the reaction involves changing all reactants to products and vice versa. When a reaction is reversed, the sign of \( \Delta H \) changes. So if after doubling the \( \Delta H \) was \( 2x \), reversing the reaction will change \( \Delta H \) to \( -2x \).
04

Combine the Changes

Both the doubling and reversing of the reaction have their individual effects on \( \Delta H \). Doubling makes \( \Delta H = 2x \) and reversing changes the sign. So, combining these transformations, the final \( \Delta H \) is \( -2x \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Reaction
When we talk about a chemical reaction, we refer to a process where substances, called reactants, are transformed into different substances, known as products. This transformation involves the breaking and forming of chemical bonds, which either absorbs or releases energy. This energy change is often expressed as a change in enthalpy, noted as \( \Delta H \), which serves as a measure of the heat absorbed or released during the reaction.
  • A negative \( \Delta H \) indicates an exothermic reaction that releases heat.
  • A positive \( \Delta H \) signals an endothermic reaction where heat is absorbed.
Understanding \( \Delta H \) is crucial because it influences how the reaction proceeds and affects its equilibrium. Whether working in a lab or analyzing chemical data, recognizing these changes can help predict reaction outcomes and their impacts on the surroundings.
Reversing Reaction
Reversing a reaction involves swapping the reactants and the products, essentially turning the reaction around. When this happens, the sign of the enthalpy change \( \Delta H \) also reverses. This is because the direction of the energy flow is switched:
  • If the forward reaction is exothermic (releases heat), the reverse reaction will be endothermic (absorbs heat), resulting in a positive enthalpy change.
  • Conversely, if the forward reaction absorbs heat, reversing it will cause it to release heat, changing \( \Delta H \) to a negative value.
By reversing a reaction, you change not only the chemicals involved but also the energetic balance, which plays a critical role in industrial applications and chemical pathways. It's akin to retracing your steps on a map, where each path covers the same distance but in the opposite direction.
Doubling Reaction
When you double a chemical reaction, you simply multiply all the amounts of reactants and products by two. This effectively scales up the reaction, which directly impacts the enthalpy change. By doing this, everything, including \( \Delta H \), doubles, since enthalpy is proportional to the amount of substance undergoing the change. For instance:
  • If the original reaction had a \( \Delta H \) of \( x \), doubling it would make \( \Delta H \) equal to \( 2x \).
  • This basic principle can be extended to scaling reactions by any factor, say tripling or halving, by simply multiplying \( \Delta H \) by the same factor.
Doubling reactions is a common technique used in chemical equations and stoichiometry to ensure the conservation of mass and energy. It's essential for calculating yields and preparing reactions at different scales, from small lab experiments to industrial processes.

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Most popular questions from this chapter

How fast (in meters per second) must an iron ball with a mass of \(56.6 \mathrm{~g}\) be traveling in order to have a kinetic energy of \(15.75 \mathrm{~J} ?\) The density of iron is \(7.87 \mathrm{~g} / \mathrm{cm}^{3} .\)

The equation for the combustion of \(2 \mathrm{~mol}\) of butane can be written $$ 2 \mathrm{C}_{4} \mathrm{H}_{10}(g)+\mathrm{O}_{2}(g) \longrightarrow 8 \mathrm{CO}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g) ; \Delta H<\mathrm{O} $$ Which of the following produces the least heat? a. Burning 1 mol of butane. b. Reacting \(1 \mathrm{~mol}\) of oxygen with excess butane. c. Burning enough butane to produce \(1 \mathrm{~mol}\) of carbon dioxide. d. Burning enough butane to produce 1 mol of water. e. All of the above reactions \((a, b, c\), and \(d)\) produce the same amount of heat.

Why is it important to give the states of the reactants and products when giving an equation for \(\Delta H ?\)

Hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\), is produced during decomposition of organic matter. When \(0.5500 \mathrm{~mol} \mathrm{H}_{2} \mathrm{~S}\) burns to produce \(\mathrm{SO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l),-309.1 \mathrm{~kJ}\) of heat is released. What is this heat in kilocalories?

Given the following (hypothetical) thermochemical equations: $$ \begin{aligned} &\mathrm{A}+\mathrm{B} \longrightarrow 2 \mathrm{C} ; \Delta H=-447 \mathrm{~kJ} \\ &\mathrm{~A}+3 \mathrm{D} \longrightarrow 2 \mathrm{E} ; \Delta H=-484 \mathrm{~kJ} \\ &2 \mathrm{D}+\mathrm{B} \longrightarrow 2 \mathrm{~F} ; \Delta H=-429 \mathrm{~kJ} \end{aligned} $$ Calculate \(\Delta H\), in \(\mathrm{kJ}\), for the equation $$ 4 \mathrm{E}+5 \mathrm{~B} \longrightarrow 4 \mathrm{C}+6 \mathrm{~F} $$

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