Rate Of A Chemical Reaction

The Wurtz reaction involves the coupling of alkyl halides in the presence of sodium metal to form higher alkanes. When ethyl iodide (C H I) and n-propyl iodide (C H I) are subjected to the Wurtz reaction, they can couple to form different hydrocarbons.
Step 1: The reaction between two ethyl iodide molecules can form butane (C H ), as shown by: The reaction between ethyl iodide and n-propyl iodide can form pentane (C H ), as shown by:
Step 2: However, propane (C H ) cannot be formed in this reaction. This is because propane would require the coupling of two propyl iodide molecules, which is not possible here, as only ethyl and n-propyl iodides are involved.
Step 3: Thus, the hydrocarbon that will not be formed is propane.

- Statement I: The octet theory primarily explains the tendency of atoms to have eight electrons in their valence shell to achieve stability. It does not directly account for the shape of molecules, which is better explained by the VSEPR (Valence Shell Electron Pair Repulsion) theory. Therefore, Statement I is not correct.
- Statement II: The octet theory does not explain the relative stability of molecules, especially in cases where molecules have expanded octets or odd-electron species.
Thus, Statement II is correct.
Final Answer: Statement-I is not correct but statement-II is correct

Step 1: Analyze the molecular geometries.
Set i: Both molecules are tetrahedral, sharing identical electron pair geometries.
Set ii: Both molecules have a pyramidal structure, albeit with bonding pairs and with lone pairs affecting the structure.
Set iii: Both are linear, having the simplest geometric arrangement.
Step 2: Comparison with Set iv. Set iv does not contain isostructural molecules due to the different spatial arrangements ( see-saw vs. square planar).

To find the equilibrium concentrations of and , we need to analyze the reaction with a given equilibrium constant ( ) of 39.0.

Initially, 1 mole of is present in a 1-litre flask at , making the initial concentration of , .

At equilibrium, let the change in concentration of that reacted be . Therefore, the equilibrium concentrations can be expressed as:

Substituting these expressions into the equilibrium expression for :

$$\frac{\mathrm{[B]}}{\mathrm{[A]}}=\; K\_c$$

Thus,

$\frac{x}{\mathrm{1-x}}=\; 39.0$

Solving for :

This implies:

• At equilibrium,

Thus, the equilibrium concentrations of and are respectively 0.025 mol L and 0.975 mol L . The correct answer is .

When sulfur reacts with concentrated sulfuric acid, sulfur undergoes oxidation and reduction. The reaction is: In this reaction:
- Sulfur (S) is oxidized to , which is the oxidized product (X).
- The sulfur (S) is also reduced to , which is the reduced product (Y).
Step 1: The sulfur reacts with sulfuric acid to form as both the oxidized and reduced product. This is because sulfur in the reaction can simultaneously undergo oxidation and reduction.
Step 2: Therefore, the correct answer is , corresponding to option (2).

The reaction describes the conversion of styrene into products via two steps:
- Step 1: The reaction of styrene with reagent X leads to the formation of compound Y. Styrene ( ) reacts with HBr (hydrobromic acid) under suitable conditions. This results in the addition of HBr across the double bond, forming phenyl ethanol ( ).
- Step 2: Upon hydrolysis and oxidation of Y, the product Z is formed. The oxidation step converts phenyl ethanol to benzaldehyde ( ). Z gives a positive 2,4-DNP test, indicating the presence of an aldehyde group, but it does not give the iodoform test, suggesting that Z does not contain a methyl ketone group.
Thus, the correct sequence is:
- X = HBr (which adds across the double bond)
- Z = benzaldehyde ( ). The correct reaction follows the sequence: Thus, the correct answer is (3).

Step 1: Understanding the reaction conditions - The ability of a metal to liberate from acid depends on its reduction potential. - A metal with a lower standard reduction potential than can displace from acid, whereas metals with a higher reduction potential cannot.
Step 2: Analysis of each reaction - Reaction I:
- Zinc has a lower reduction potential (-0.76 V) than , so it can displace . - gas is liberated. - Reaction II: No reaction.
- Copper has a higher reduction potential (+0.34 V) than , so it does not react with HCl. - gas is not liberated. - Reaction III: NO gas instead of .
- acts as an oxidizing agent, reducing to . - gas is not liberated.
Final Answer: Reactions II and III do not liberate .

Step 1: Understanding the Given Reaction - The reaction provided is: - Here, 1 mole of Xenon reacts with 2 moles of Fluorine gas to form XeFsubscript{4}.
Step 2: Analyzing the Ratio - The ratio of Xe to Fsubscript{2} in the equation is 1:2.
- However, in an extended fluorination process, more fluorine gas may be required to sustain and complete the reaction.
Step 3: Applying Stoichiometry - From experimental data, the reaction sometimes requires 1:5 (Xe:Fsubscript{2}) under certain conditions.
- This is because excess fluorine helps drive the reaction forward.

Step 1: Understanding the Rate of Reaction - The rate of reaction is defined using the rate of disappearance of reactants or the rate of formation of products. - The general expression for rate is: where is the stoichiometric coefficient of species .
Step 2: Applying the Rate Expression - Given: Since the balanced equation gives a stoichiometric coefficient of 5 for : Final Answer: The correct rate of the reaction is .

The decomposition of is a first-order reaction: Let the initial pressure of be atm. After time , let the pressure of that has decomposed be atm.
The total pressure at time is given as atm. The change in pressure due to the reaction is: Thus, the total pressure at time is: Solving for : The pressure of at time is: For a first-order reaction, the rate constant is given by: considering the given options and the context, the correct answer is: Final Answer:

The wavenumber of the first line of the Lyman series of hydrogen is given by the Rydberg formula for hydrogen: The second line of the Balmer series for the ion is also given by the Rydberg formula: Since the wavenumbers are equal, we can equate the Rydberg constants for both: After solving for , we find that the ion corresponding to this condition is .
Final Answer: .