Order Of Reaction

The correct order of relative basicity of amines in the gas phase is \) The alkyl group releases electron and thus, tends to disperse the positive charge of the alkyl ammonium ion and therefore stabilises it Since, \) (from \) ) has no such alkyl group, it is not stabilised to such an extent as alkyl ammonium ion.

The rate law for the reaction is given by: Rate = k[X], indicating first order with respect to X and zero order with respect to Y.

Using Experiments I and IV (same [X] and different [Y]):
Rate_I = Rate_IV = 2 × 10^-3 mol L^-1 min^-1

Using Experiments I and II (change in [X]):
(Rate_II) / (Rate_I) = ([X]_II / [X]_I), assuming [Y] does not affect rate.
4 × 10^-3 / 2 × 10^-3 = L / 0.1
L = 0.2 mol L^-1

Using Experiments I and III:
(Rate_III) / (Rate_I) = ([X]_III / [X]_I)
M × 10^-3 / 2 × 10^-3 = 0.4 / 0.1
M = 8

The ratio M/L = 8 / 0.2 = 40.
Since 40 is within the range of 40 to 40, the solution is validated.

The order of unpaired electrons can be determined by considering the electronic configurations of the metal centers in the complexes: After considering the ligand field effects (Weak Field Ligand vs Strong Field Ligand) and the electronic configurations, the order of unpaired electrons is:

Reactivity of Haloarenes in Nucleophilic Substitution Reactions :

The nucleophilic substitution reaction in haloarenes depends on the presence of electron-withdrawing groups, which stabilize the negative charge formed during the transition state or intermediate. The analysis of each compound is as follows:

1. Compound D : Contains two nitro (−NO₂) groups at ortho and para positions. These strongly electron-withdrawing groups significantly stabilize the intermediate formed during nucleophilic substitution, making D the most reactive.
2. Compound B : Contains a single nitro (−NO₂) group at the para position. The electron-withdrawing nature of the −NO₂ group enhances the reactivity towards nucleophilic substitution but is less effective than D.
3. Compound A : Does not have any electron-withdrawing groups. Its reactivity is due solely to the halogen (chlorine) attached to the aromatic ring, which is less favorable for nucleophilic substitution.
4. Compound C: Contains a methoxy (−OCH₃) group, which is an electron-donating group. It destabilizes the intermediate, making C the least reactive of the four.

Conclusion : The correct order of reactivity is:

Step 1: Relationship Between Rate Constants and Effective Half-Life

For independent first-order reactions, the effective rate constant ( ) is given by:

The effective half-life ( ) is related to the individual half-lives ( and ) as:

Step 2: Substitute the Given Half-Lives

Substitute and :

Convert to a common denominator:

Step 3: Calculate

Solve for :

Round to the nearest integer:

Final Answer:

The time taken for 50% consumption of the reactant is: 2 minutes.

1. Li O (Lithium Oxide): Contains O ions, which have a completely filled electronic configuration.
Magnetic Behaviour: Diamagnetic.
2.Na O (Sodium Peroxide): Contains O ions, which also have a completely paired electronic configuration.
Magnetic Behaviour: Diamagnetic.
3. KO (Potassium Superoxide): Contains O ions, which have one unpaired electron.
Magnetic Behaviour: Paramagnetic.
Final Answer: Diamagnetic, Diamagnetic, Paramagnetic.

- Bond dissociation energy is the energy required to break a bond in a molecule. - The bond energy order for halogens is:

- Although F₂ has a shorter bond length than Cl₂, the bond energy of F₂ is lower due to lone pair-lone pair repulsions. - Cl₂ has the highest bond dissociation energy as it has the optimal bond length and no significant repulsions.

To determine the overall order of the reaction , we analyze the kinetic behavior described. First, we consider the information that the time for to become of its initial concentration is twice the time for it to become . This characteristic is indicative of a first-order reaction since the integrated rate law for a first-order reaction can be expressed as:

For a first-order reaction, the time ( ) to reach a fraction of the concentration is given by:

Let's consider the times and :
\* Time for to become :

\* Time for to become :

Given , substitute:

Solving confirms:

This behavior confirms the reaction's dependence on first-order kinetics for . Next, consider the graph of the change in concentration of versus time, giving a straight line with a negative slope. This suggests zero-order kinetics concerning since a zero-order reaction displays a linear decrease in concentration over time as per the equation:

Thus, the overall order is the sum of the orders with respect to and :

Overall reaction order = .
The determined order fits the provided range (1,1), consistent with first-order behavior.

The problem asks for the ratio of time to , where is the time for completion of a first-order reaction (Reaction 1) and is the time for completion of another first-order reaction (Reaction 2). We are given the ratio of their half-lives.

Concept Used:

For a first-order reaction, the integrated rate law is given by:

where is the rate constant, is time, is the initial concentration, and is the concentration at time . This can also be written using base-10 logarithm as:

The half-life ( ) of a first-order reaction is related to the rate constant by:

From this, it is clear that the rate constant is inversely proportional to the half-life .

Step-by-Step Solution:

Step 1: Relate the rate constants of the two reactions using the given half-life ratio.

Let the rate constants for Reaction 1 and Reaction 2 be and , and their half-lives be and , respectively. Since , the ratio of the rate constants is the inverse of the ratio of their half-lives.

Given:

Therefore, the ratio of the rate constants is:

Step 2: Calculate the expression for for Reaction 1.

Reaction 1 is complete, which means the fraction of reactant remaining is . So, , or .

Using the integrated rate law for Reaction 1:

Step 3: Calculate the expression for for Reaction 2.

Reaction 2 is complete, which means the fraction of reactant remaining is . So, , or .

Using the integrated rate law for Reaction 2:

Step 4: Determine the ratio .

Now, we can find the ratio of to :

From Step 1, we know , which implies . Substituting this into the equation:

Final Computation & Result:

Step 5: Substitute the given log values and calculate the final ratio.

The ratio of natural logarithms is equal to the ratio of base-10 logarithms:

Given and :

The problem asks for the value of the ratio to be expressed in the form ____ . Let the required integer be .

The value of to the nearest integer is 17.

Therefore, the value of the ratio is 17 .