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The energy released by the reaction \(\mathrm{A} \rightarrow \mathrm{B}\) is \(400 \mathrm{~kJ} / \mathrm{mol}\). (a) What is \(\Delta E_{\text {forward } \mathrm{rxn}}\) ? (b) What is \(\Delta E_{\text {reverse rxn }}\) ?

Short Answer

Expert verified
(a) The change in energy, ΔE, for the forward reaction is -400 kJ/mol. (b) The change in energy, ΔE, for the reverse reaction is 400 kJ/mol.

Step by step solution

01

Part (a): Find ΔE for the forward reaction

For the forward reaction A → B, by definition, the energy released is given as negative of the change in energy ΔE. That is, -ΔE = Energy released. We have the energy released as 400 kJ/mol. So, -ΔE = 400 kJ/mol. Now we can find ΔE as: ΔE = -400 kJ/mol Thus, the change in energy, ΔE, for the forward reaction is -400 kJ/mol.
02

Part (b): Find ΔE for the reverse reaction

The reverse reaction can be represented as B → A. Since the energy released in the forward reaction A → B is 400 kJ/mol, in the reverse reaction B → A the energy is absorbed instead. This means the change in energy ΔE for the reverse reaction is positive, and equal in magnitude to that of the forward reaction. Thus, the change in energy, ΔE, for the reverse reaction is 400 kJ/mol.

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Key Concepts

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

Forward Reaction
In chemistry, the forward reaction is when reactants convert into products. In our exercise, the chemical reaction is represented as \( \mathrm{A} \rightarrow \mathrm{B} \). This indicates that substance \( A \) transforms into substance \( B \). During this transformation, there is a release of energy. This release is energy that was stored in the chemical bonds of \( A \). For our reaction, this energy is given as \( 400 \text{ kJ/mol} \). This means that for every mole of \( A \) that reacts, \( 400 \text{ kJ} \) of energy is released.

Notably, in thermodynamics, whenever energy is released during a reaction, it is considered to be lost from the system's perspective. To account for this, the energy change for a forward reaction is expressed as a negative value. So, the energy change \( \Delta E_{\text{forward rxn}} \) is \(-400 \text{ kJ/mol} \).

The negative sign is crucial because it indicates an exothermic reaction, a term used when energy leaves the system. Exothermic reactions are common in everyday life, including combustion, like burning wood or fuel.
Reverse Reaction
The reverse reaction is simply when the product converts back to the reactant. For our specific example, it's \( \mathrm{B} \rightarrow \mathrm{A} \). This reaction is the exact opposite of the forward reaction. Here, instead of releasing energy, the system absorbs it.

Since energy was released in making \( B \) from \( A \), breaking \( B \) back into \( A \) requires energy input. This is equivalent to the energy initially released, \( 400 \text{ kJ/mol} \). Therefore, for the reverse reaction, the energy change, \( \Delta E_{\text{reverse rxn}} \), is \(+400 \text{ kJ/mol} \).

The positive sign represents that the reaction is endothermic, a term for reactions or processes that absorb energy from the surroundings. Examples of endothermic processes include photosynthesis in plants and melting ice.
Energy Change Calculation
Calculating energy changes, \( \Delta E \), in reactions is a straightforward process but essential in understanding thermodynamics. These values help us understand whether a reaction is endothermic or exothermic and by how much energy.

Let's break it down:
  • For the forward reaction \( \mathrm{A} \rightarrow \mathrm{B} \), we take the given energy release value and assign it a negative sign because energy is leaving the system, which gives us \(-400 \text{ kJ/mol} \).
  • For the reverse reaction \( \mathrm{B} \rightarrow \mathrm{A} \), we know this is the reverse of the forward reaction. Thus, the same amount of energy must be absorbed. We use a positive sign to indicate an energy gain, leading to \(+400 \text{ kJ/mol} \).

In summary, understanding \( \Delta E \) allows us to know both the direction and the magnitude of energy flow involved in chemical reactions.

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Most popular questions from this chapter

Determine the value of \(E_{\mathrm{a}}\) and \(\Delta E_{\mathrm{rxn}}\) for each case below. Also indicate whether each reaction is endothermic or exothermic. $$\begin{array}{lccc} & \begin{array}{l} \text { Energy of } \\ \text { reactants, kJ } \end{array} & \begin{array}{l} \text { Energy of } \\ \text { transition state, kJ } \end{array} & \begin{array}{l} \text { Energy of } \\ \text { products, kJ } \end{array} \\ \hline \text { (a) } & 100 & 150 & 130 \\ \text { (b) } & 100 & 150 & 70 \\ \text { (c) } & 50 & 175 & 130 \\ \text { (d) } & 20 & 40 & 10 \\ \hline \end{array}$$

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A reaction \(\mathrm{A}+\mathrm{B} \rightarrow\) Product is run in a balloon. (Both A and B are gases.) The balloon has a volume of 1 L and is initially loaded with 1 mole of \(\mathrm{A}\) and \(1 \mathrm{~mole}\) of \(\mathrm{B}\). The reaction has the rate law Rate \(=k[\mathrm{~A}]\) The reaction is run again using the same amount of reactants, but this time in a balloon that has a volume of \(0.5 \mathrm{~L}\). How much faster will the reaction proceed in the smaller balloon? Explain your answer.

Which of the following are substitution reactions: (a) \(\mathrm{CH}_{3} \mathrm{Br}+\mathrm{I}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{I}+\mathrm{Br}^{-}\) (b) \(\mathrm{H}_{2}+\mathrm{Br}_{2} \rightarrow 2 \mathrm{HBr}\) (c) \(\mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{H}_{2} \rightarrow \mathrm{CH}_{3}-\mathrm{CH}_{3}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}+\mathrm{OH}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{Cl}^{-}\)

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