Problem 29
Question
The standard state Gibb's free energy change for the i. cis-2-pentene = trans-2-pentene is \(-3.67 \mathrm{~kJ} \mathrm{~mol}\) at \(400 \mathrm{~K}\). If more trans-2-pentene is added to the reaction vessel (a) more cis-2-pentene is formed (b) equilibrium shifts in the forward direction (c) equilibrium remains unaltered (d) more trans-2-pentene is produced
Step-by-Step Solution
Verified Answer
(a) More cis-2-pentene is formed.
1Step 1: Identify the Reaction Direction
The given standard Gibbs free energy change for the reaction \( \text{cis-2-pentene} \rightarrow \text{trans-2-pentene} \) is \( \Delta G^\circ = -3.67 \text{ kJ/mol} \). Since \( \Delta G^\circ \) is negative, this indicates that at standard conditions, the reaction favors the formation of trans-2-pentene (forward direction).
2Step 2: Apply Le Chatelier's Principle
According to Le Chatelier's Principle, if more trans-2-pentene is added to the system at equilibrium, the system will shift to counteract this change, which means the equilibrium will shift towards the reverse reaction, forming more cis-2-pentene.
3Step 3: Determine the Correct Option
Given that adding more trans-2-pentene will shift the equilibrium towards the formation of cis-2-pentene, option (a) more cis-2-pentene is formed, is the correct choice.
Key Concepts
Le Chatelier's PrincipleReaction EquilibriumThermodynamicsStandard State Conditions
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in chemistry that explains how a system at equilibrium responds to changes in concentration, temperature, or pressure. This principle helps us predict the direction in which a reaction will shift to restore equilibrium.
When a change occurs in a system at equilibrium, such as adding more reactants or products:
When a change occurs in a system at equilibrium, such as adding more reactants or products:
- The system will shift in the direction that opposes this change.
- If more products are added, the equilibrium will shift towards the reactants.
- If more reactants are added, the equilibrium will shift towards the products.
Reaction Equilibrium
Reaction equilibrium refers to the state where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time.
In a chemical equation like cis-2-pentene \( \leftrightarrow \) trans-2-pentene:
In a chemical equation like cis-2-pentene \( \leftrightarrow \) trans-2-pentene:
- The reaction has reached equilibrium when the amounts of cis-2-pentene and trans-2-pentene stay steady.
- No further net change in concentrations will occur unless the system is disturbed.
Thermodynamics
Thermodynamics is the study of energy and its transformations. In the realm of chemical reactions, it explores concepts like Gibbs free energy, enthalpy, and entropy.
Gibbs Free Energy
Gibbs free energy \( (\Delta G) \) is used to determine the spontaneity of a reaction:
Gibbs Free Energy
Gibbs free energy \( (\Delta G) \) is used to determine the spontaneity of a reaction:
- If \( \Delta G \) is negative, the reaction is spontaneous under the given conditions.
- If \( \Delta G \) is positive, the reaction is non-spontaneous.
Standard State Conditions
Standard state conditions are a set of established conventions in thermodynamics to provide a reference point for measuring properties of substances.
Under standard state conditions:
Under standard state conditions:
- The pressure is typically \( 1 \text{ atm} \).
- The temperature is usually \( 298.15 \text{ K} \) (or \( 25 ^\circ \text{C} \)).
- Solutions are at a concentration of \( 1 \text{ M} \).
Other exercises in this chapter
Problem 25
In the manufacture of ammonia by haber's process $$ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{3}(\mathrm{~g})=2 \mathrm{NH}_{3}(\mathrm{~g})+92.3 \mathrm{~kJ} $
View solution Problem 26
\(\mathrm{CaCO}_{3}=\mathrm{CaO}+\mathrm{CO}_{2}\) reaction in a lime kiln goes to completion because (a) \(\mathrm{CaO}\) does not react with \(\mathrm{CO}_{2}
View solution Problem 24
In the manufacture of ammonia by haber's process $$ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{3}(\mathrm{~g})=2 \mathrm{NH}_{3}(\mathrm{~g})+92.3 \mathrm{~kJ} $
View solution