Chemical Kinetics

The correct answer is 75



Therefore, percentage of original activity remaining after 83 days is 75%.

The correct answer is 59
T₁ = 300 K (Rate constant)
K₂ = 2K₁, on increase temperature by 9K
T₂ = 309 K
Ea = ?



= 58988.1 J/mole ≃ 59 kJ/mole

The correct answer is 216


E_a = 26000 × 8.314
= 216164 J
= 216 kJ

Calculation of Time for Equal Concentrations of A and B:

The decay of a substance follows the formula:

N(t) = N₀ × (1/2)^t/t_1/2

Where:

• N(t): Concentration at time t
• N₀: Initial concentration
• t_1/2: Half-life of the substance

Let the initial concentration of A be N_A, and the initial concentration of B be N_B = 4N_A.

1. For substance A:
   N_A(t) = N_A × (1/2)^t/15
2. For substance B:
   N_B(t) = 4N_A × (1/2)^t/5
3. Set the concentrations equal:
   N_A × (1/2)^t/15 = 4N_A × (1/2)^t/5
4. Cancel N_A from both sides:
   (1/2)^t/15 = 4 × (1/2)^t/5
5. Rewrite 4 as 2²:
   (1/2)^t/15 = (1/2)^{t/5 - 2}
6. Equating the exponents:
   t/15 = t/5 - 2
7. Solve for t:
   t/15 - t/5 = -2
   Multiply through by 15 to eliminate the fractions:
   t - 3t = -30
   -2t = -30
   t = 15 minutes

Conclusion: It will take 15 minutes for the concentrations of A and B to become the same.

Vanderwaal and torsional strain. Hence it must be most stable.

So, the correct option is (A).

Correct answer is (d)

Step 1: Analyze Statement I
Sulfanilic acid contains an amine group ( ) which is attached to the benzene ring, and a sulfonic acid group ( ). It does not contain a carboxyl group ( ). Esterification is a characteristic reaction of carboxylic acids. Therefore, Statement I is incorrect.
Step 2: Analyze Statement II
Lassaigne's test is used to detect the presence of nitrogen, sulfur, halogens, and phosphorus in organic compounds. Sulfanilic acid contains sulfur and nitrogen. The red color in Lassaigne's test is due to the formation of ferric thiocyanate when sulfur is present. Thus, Statement II is correct.

The correct answer is 3

A:k=AA−RTEa​
As Ea increases k decreases
B : Temperature coefficient =kT​kT+10​​
Option (C) is wrong. Δk may be greater or lesser depending on temperature.
D:lnk=lnA−RTEa​

Ne > Cl₂ > UF₆

Step 1: Reaction and First-Order Kinetics Formula
The reaction is:

At , the concentration of A is . At , the concentration of A reduces to . For first-order reactions:

Step 2: Substitute the Values
Substitute , , and :


Step 3: Calculate the Time
Substitute the values:



Conclusion: The time taken by A to reduce from 7 g to 2 g is .

The correct answer is 3.
At 87.5% consumption, we have 3 half lives.

Step 1: Equations and Their Enthalpy Changes

We are given:

The target equation is:

Step 2: Manipulate the Given Equations

Reverse and divide equation (1) by 2:

Keep equation (2) as is:

Step 3: Add the Modified Equations

Combine the two equations:

Cancel out :

Step 4: Calculate the Enthalpy Change

Add the enthalpy changes of the modified equations:

Reorganize the terms:

Final Answer:

The enthalpy change for the reaction is:

For a reaction, the half-life is related to the initial pressure ( ) by:
where is the order of the reaction.
Taking the ratio of half-lives at different pressures:

Using the given data:

Substitute and :

Simplify:

Taking the logarithm:

Thus, the order of the reaction is .

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

where is the concentration of at time , is the initial concentration of , and is the rate constant.
The half-life ( ) of a zero-order reaction is given by:

Given that , we can find the rate constant :

We are asked to find the time taken for the concentration of to reduce to one-fourth of its initial value. Let this time be . So, . Substituting this into the integrated rate law:


Substituting the value of we found earlier:

Simplify:

Final Answer:
The time taken is .

Rate of reaction is given by the change in concentration of FeSO₄: Substitute the given values: From the given reaction, the rate of production of Fe₂(SO₄)₃ is related to the rate of FeSO₄: Substitute the value of : Thus, the rate of production of Fe₂(SO₄)₃ is mol L⁻¹ s⁻¹. Hence,

Given: The reaction is first order with respect to both and . The rate law is:

where:

• : Rate of the reaction
• : Rate constant
• and : Concentrations of and , respectively

Step 1: Set Up Equations

Using the given data, we can write the following equations:

1. — (1)
2. — (2)
3. — (3)

Step 2: Solve for

Divide equation (2) by equation (1):

Simplify:

Step 3: Solve for

Divide equation (3) by equation (1):

Simplify:

Final Answer:

The values of and are 80 and 2, respectively.

The problem asks to determine the ratio of the equilibrium constant in terms of partial pressure ( ) to the equilibrium constant in terms of molar concentration ( ) for the reaction: .

Concept Used:

The relationship between and for a reversible gaseous reaction is given by the equation:

where:

is the ideal gas constant.

is the absolute temperature in Kelvin.

is the change in the number of moles of gaseous components, calculated as the difference between the sum of the stoichiometric coefficients of the gaseous products and the sum of the stoichiometric coefficients of the gaseous reactants.

Step-by-Step Solution:

Step 1: Identify the stoichiometric coefficients of the gaseous reactants and products from the given balanced equation.

The reaction is:

The stoichiometric coefficient of the gaseous product ( ) is 1.

The sum of the stoichiometric coefficients of the gaseous reactants ( and ) is .

Step 2: Calculate the value of .

Using the formula for :

Step 3: Substitute the value of into the relationship .

Step 4: Rearrange the equation to find the required ratio .

Final Computation & Result:

The expression can be simplified as follows:

Therefore, the ratio for the given reaction is . This corresponds to option (4).