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For the decomposition of one mole of nitrosyl chloride, \(\Delta H=38 \mathrm{~kJ}\). $$ \mathrm{NOCl}(g) \longrightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g) $$ The activation energy for this reaction is \(100 \mathrm{~kJ}\) a, Is this reaction exothermic or endothermic? b. What is the activation energy for the reverse reaction? c. If a catalyst were added to the reaction, how would this affect the activation energy?

Short Answer

Expert verified
a. Endothermic; b. 138 kJ; c. Lowers activation energy.

Step by step solution

01

Understand the Reaction

The given reaction is \( \mathrm{NOCl}(g) \longrightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g) \) with \( \Delta H=38 \mathrm{~kJ} \). The \( \Delta H \) represents the enthalpy change of the reaction.
02

Determine Reaction Type (Exothermic vs Endothermic)

Since \( \Delta H \) is given as \( 38 \mathrm{~kJ} \) and is positive, this indicates that the reaction absorbs heat. Thus, the reaction is endothermic (energy is absorbed).
03

Activation Energy for Reverse Reaction

For the reverse reaction, the activation energy \( E_{a,reverse} \) is derived by adding the activation energy of the forward reaction to \( \Delta H \) because \( E_{a,forward} - \Delta H = E_{a,reverse} \). Thus, \( E_{a,reverse} = 100 \mathrm{~kJ} + 38 \mathrm{~kJ} = 138 \mathrm{~kJ} \).
04

Effect of Catalyst on Activation Energy

A catalyst provides an alternative pathway with a lower activation energy for the reaction. Thus, if a catalyst were added to the reaction, it would decrease the activation energy for both the forward and reverse reactions.

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Key Concepts

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

Endothermic Reactions
In thermochemistry, endothermic reactions are processes in which energy is absorbed from the surroundings. This absorption is typically in the form of heat. Such reactions have a positive enthalpy change, which is denoted as \( \Delta H \). A positive \( \Delta H \) means that the products have higher energy content compared to the reactants.
In the reaction provided, the decomposition of nitrosyl chloride (NOCl) into nitric oxide (NO) and chlorine gas (Clâ‚‚), the \( \Delta H = 38 \, \mathrm{kJ} \/ \mathrm{mol} \), signifies an endothermic process.
- This indicates that the system requires 38 kJ of energy per mole for the reaction to proceed.
- The absorption of heat from the surroundings causes the solution or environment to feel cooler.
Understanding whether a reaction is endothermic helps predict the behavior of the system when conditions change. For example, increasing the temperature will generally drive an endothermic reaction forward, as it compensates for the absorbed energy.
Activation Energy
Activation energy (\( E_{a} \)) is the energy barrier that must be overcome for a chemical reaction to proceed. It is the minimum quantity of energy needed to convert reactants into products. Without sufficient energy for this activation, the reaction will not occur.
For the decomposition of NOCl, the given activation energy is \( 100 \, \mathrm{kJ} \/ \mathrm{mol} \) for the forward reaction.
  • This energy helps break the bonds in NOCl, allowing the formation of NO and Clâ‚‚.
  • High activation energy indicates that the reaction requires more energy input to occur.
The activation energy for the reverse reaction is calculated by adding the forward reaction's activation energy to the enthalpy change: \( E_{a,reverse} = 100 \, \mathrm{kJ} + 38 \, \mathrm{kJ} = 138 \, \mathrm{kJ} \).
This calculation shows that the reverse reaction has an even higher energy barrier to surpass. Knowing these values helps chemists modify reaction conditions to favor desired pathways.
Catalysis
Catalysis is a process where a substance, known as a catalyst, is used to speed up a chemical reaction without being consumed in the reaction itself. Catalysts work by providing an alternative pathway with a lower activation energy, thereby increasing the reaction rate.
In the provided exercise, adding a catalyst to the decomposition of NOCl would lower the activation energy of the reaction.
  • This results in easier bond breaking due to lower energy threshold.
  • The catalyst affects both the forward and reverse reactions, lowering the activation energies symmetrically.
By lowering the activation energy, substances that act as catalysts allow reactions to proceed faster or at lower temperatures. Thus, understanding and using catalysis is crucial in industrial processes to enhance efficiency and reduce costs.

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Most popular questions from this chapter

Ammonium nitrite, \(\mathrm{NH}_{4} \mathrm{NO}_{2},\) decomposes in solution, as shown here. $$ \mathrm{NH}_{4} \mathrm{NO}_{2}(a q) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(I) $$ The concentration of \(\mathrm{NH}_{4}^{+}\) ion at the beginning of an experiment was \(0.500 \mathrm{M}\). After 3.00 hours, it was \(0.432 \mathrm{M}\). What is the average rate of decomposition of \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) in this time interval?

The reaction of water with \(\mathrm{CH}_{3} \mathrm{Cl}\) in acetone as a solvent is represented by the equation $$ \mathrm{CH}_{3} \mathrm{Cl}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{CH}_{3} \mathrm{OH}+\mathrm{HCl} $$ The rate of the reaction doubles when the concentration of \(\mathrm{CH}_{3} \mathrm{Cl}\) is doubled and it quadruples when the concentration of \(\mathrm{H}_{2} \mathrm{O}\) is doubled. a. What is the unit for \(k ?\) b. Calculate \(k\) if \(\mathrm{CH}_{3} \mathrm{OH}\) is formed at a rate of \(1.50 \mathrm{M} / \mathrm{s}\) when \(\left[\mathrm{CH}_{3} \mathrm{Cl}\right]=\left[\mathrm{H}_{2} \mathrm{O}\right]=0.40 \mathrm{M}\)

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