Chemical Kinetics

Kinetic energy of
and




in

The reaction is A 3B. The rate law is given as rate = k[A] . This means it is a zero-order reaction with respect to A. For a zero-order reaction, the rate is constant and independent of the concentration of A: rate = . The integrated rate law for a zero-order reaction is: where is the concentration of A at time , is the initial concentration of A, and is the rate constant. From the given data: At , . At , . We can use this information to find the rate constant . . . . . Now, we need to find the concentration of A at , which is given as . This is the amount of A reacted. . . . . So, , which means . This is the decrease in concentration of A, which is the amount of A reacted. The reaction is A 3B. For every 1 mole of A that reacts, 3 moles of B are formed. So, the concentration of product B formed, , is . Amount of A reacted by time is . . Concentration of B formed at : . . Let's check the options: % Option (a) . This is . % Option (b) . % Option (c) . This is very close to . % Option (d) . The calculated concentration of product B is . Option (c) is . The slight difference might be due to rounding or if "0.1-x" implies itself is the answer for something. "What is the concentration (in mol L ) of the product after 20 s". This means . My value is . Option (c) is . It is likely the intended answer through some rounding of or the values. Amount of A reacted . Conc. of B . Let's check if in refers to something else. The table gives "Concentration of the reactant (mol L )" At , it is . This is not defined further in the question context. However, usually represents the amount reacted from initial. So . Here . So . The problem wrote for . This suggests is the amount of A reacted by . My calculation Then . It's possible the options are constructed with more precision in . . This is exact. Amount of A reacted in 20s . Amount of B formed . The value is or . . . They are close. is . Perhaps there is a error somewhere or a slightly different . If initial conc. was , and . If values were like ; . Then . Then . . This matches option (c). This means the value at s might imply some precision limit. If (from ), then change in 15s. The data in implies . This is exact. My calculation of is robust from the given numbers. Option (c) is likely what the question setter derived due to using an approximation for like or for . Rate constant . Amount of A reacted at is . Concentration of B formed is . The closest option is (c) . The discrepancy is . This is likely a precision/rounding issue in the question's intended solution path.

For a first order reaction, time for 93.75% completion = 4 half-lives ( )
So,

For a zero order reaction, the integrated rate law is: Rearranged:
Substitute: mol/L, mol/L, mol L s
s

This is a pseudo first-order reaction because water is in large excess. Use: Wait—99% hydrolysis implies 1% remains: BUT—in the question, rate constant is based on base-10 log for remaining fraction: So, final answer: However, if we assume the intended answer is based on only 90% hydrolyzed, then: Still results in: So, based on correct interpretation, answer should be (not 0.2303) but: % Correction Note Note: The answer marked as correct in the key (option 1) corresponds to 90% hydrolysis rather than 99%. Correct value should be 0.4606.

- A at body center = 1 atom per unit cell (fully inside) - B at 8 corners = atom - C at 3 face centers (half of 6) = atoms To remove decimals, multiply all by 2: Empirical formula = A B C

Using the Arrhenius equation: , where at , at , and . First, calculate: Substitute: This matches option 2. Thus, the correct answer is 33.45.

Step 1: Write the balanced redox reaction.
Step 2: Identify the products of the reaction.
The products are , , and . Step 3: Understand autocatalysis.
Autocatalysis is when a product of the reaction catalyzes the reaction itself. Step 4: Determine which product acts as a catalyst.
ions act as a catalyst for this reaction. The reaction is slow initially and speeds up as is formed. Thus, the autocatalyst is .

For a first-order reaction, the relationship between the concentration at any time and the initial concentration is given by: where: - is the initial concentration, - is the concentration at time , - is the rate constant. We are given: - , - , - . The time is given by the formula: Substitute the values into the equation: Thus, the correct answer is .

The enthalpy change for the reaction is given as: For the reaction where 2 moles of are formed, we divide the total enthalpy change by 2 to find the enthalpy of formation for one mole of : Thus, the standard enthalpy of formation of is .

Given:

• Mass of H SO = 4.9 g
• Volume of solution = 250 mL = 0.25 L
• Molar mass of H SO = 98 g/mol

Moles of H SO : Molarity :

The equilibrium constant is given by the equation: We are given: Substituting the values into the equilibrium constant expression: Thus, the rate of the backward reaction is .

For a zero order reaction, the rate of reaction is independent of the concentration of reactants. The rate law for the reaction can be written as: Given that the rate constant , the rate of hydrogen formation is given by: Thus, the rate of hydrogen formation is . The correct answer is option (3).

For a first-order reaction, the rate constant can be found using the formula: where is the initial concentration, and is the concentration at time . For 40% completion, 60% of the reactant remains: Given , , and , so: For 40% completion, : Now, for 80% completion, 20% of the reactant remains: Given , . The time for 80% completion is: This is closest to 157.8 min.
So, the time required for 80% completion is 157.8 min.

Let's evaluate the given statements:
- Assertion (A): A catalyst generally increases the rate of a reaction. This is correct. A catalyst increases the rate of a reaction by providing an alternative reaction pathway, allowing the reaction to proceed faster.
- Reason (R): It lowers the activation energy of a reaction by providing a new path. This is also correct. The primary mechanism by which a catalyst increases the reaction rate is by lowering the activation energy, which is the energy barrier that reactants must overcome to form products.
Now, we need to determine if (R) is the correct explanation of (A). Since a catalyst increases the reaction rate precisely because it lowers the activation energy (as stated in R), the reason directly explains the assertion. Therefore, both (A) and (R) are correct, and (R) is the correct explanation of (A).
So, the correct answer is Both (A) and (R) are correct and (R) is the correct explanation of (A).

Rate law:
From experiment 1 and 2: From experiment 1 and 3:

Step 1: Write the rate expression for the given reaction.
For the reaction 3X(g) 2Y(g) + Z(g), the rate of reaction can be expressed in terms of the rate of disappearance of the reactant and the rate of formation of the products:
$ \frac{d[X]}{dt} \frac{d[Y]}{dt} \frac{d[Z]}{dt} -\frac{d[X]}{dt} = 7.2 \times 10^{-3} ^{-1} ^{-1} \frac{d[Y]}{dt} -\frac{d[X]}{dt} $ Final Answer:

- Statement I is correct: order of reaction is the sum of powers of concentration in the rate law.
- Statement II is incorrect: it can be fractional or zero.
- Statement III is correct: order is found by experimental data, not by reaction stoichiometry.
- Statement IV is incorrect: it is applicable to all reactions, though mechanisms might be complex.

Step 1: Use the integrated rate law for zero order reactions
Step 2: For 100% completion,

- For a first-order reaction, the half-life is given by:
- Given , :
- This matches option (B).

Step 1: Compare with the Arrhenius equation. Arrhenius equation for a first-order reaction: Step 2: Match given equation. Given: So comparing: Hence, activation energy So

Step 1: Understand Zero-Order Reactions.
For a zero-order reaction, the rate of the reaction is independent of the concentration of the reactants. The rate law for a zero-order reaction is given by: where is the rate constant. Step 2: Apply the given information.
The problem states that the decomposition of AB (g) is a zero-order reaction. The rate constant, , at 300 K is given as mol L s .
Since the reaction is zero-order, the rate of the reaction is equal to the rate constant, regardless of the concentration of AB . Step 3: Determine the rate of reaction.
The rate of the reaction is equal to the rate constant , which is mol L s . The concentration of AB is irrelevant for a zero-order reaction. Step 4: Match the rate with the options.
The rate of the reaction is mol L s , which corresponds to option (2).

• Formula for Half-Life of First-Order Reaction: For a first-order reaction, the half-life t_1/2 is given by t_1/2 = ln 2 / k, where k is the rate constant and ln 2 ≈ 0.693.
• Given Value: k = 0.0693 min^-1.
• Calculate Half-Life:
  t_1/2 = 0.693 / 0.0693 ≈ 10 min.
• Option Analysis:
  • (1) 10 min: Matches the calculated value. This is correct.
  • (2) 15 min: Too high.
  • (3) 5 min: Too low.
  • (4) 20 min: Too high.
• Conclusion: The half-life is (1) 10 min.

For a first-order reaction, the integrated rate law is: $ [A]_t t [A]_0 k [A]_t = \frac{1}{2}[A]_0 t_{1/2} \frac{3}{4} [A]_t = \frac{1}{4}[A]_0 t_{3/4} $

For first order reaction: Converting to log base 10:

The average velocity of a reaction is: