Chemical Kinetics

First-order reaction kinetics:

For the reaction A → Products with first-order kinetics:

Rate law:

Integrated form: or

Expected variations for first-order kinetics:

(A) [A] vs t:

• Should show exponential decay curve
• Correct representation

(B) -d[A]/dt vs [A]:

• From the rate law:
• This is a direct proportionality
• Should show a straight line through the origin with slope k
• If the graph shows anything other than a straight line (such as a curved line), it is INCORRECT

(C) -d[A]/dt vs t:

• Since rate = k[A] and [A] decreases exponentially with time
• The rate also decreases exponentially
• Should show exponential decay curve
• Correct representation

(D) log[A] vs t:

• From:
• Should show a straight line with negative slope
• Correct representation

Identifying the incorrect plot:

For first-order kinetics, the plot of versus must be a straight line through the origin, since the rate is directly proportional to concentration.

If graph (B) shows a curved line instead of a straight line, it represents an INCORRECT variation.

Answer: (B)

Zero-order reaction

For a zero-order reaction:

Fraction completed = .

Given:

50% completion time .

So at 50% completion:

Thus,

For 80% completion

80% completion means:

Use zero-order equation:

Substitute :

Cancel :

Step 1: Understanding the decay process.
The decay of a substance follows first-order kinetics, where the decay constant is related to the half-life ( ) by the equation: The decay of reaches one-fourth of its initial amount in 29.8 hours, which corresponds to 2 half-lives. Thus, we have:

Step 2: Calculating the decay constant.
Using the equation for half-life, we can solve for :

Step 3: Conclusion.
The decay constant is .

The growth of the bacterial colony follows an exponential pattern since each bacterium divides to form two daughter cells every 20 minutes. We aim to determine the time required for the concentration to increase from its initial value to a harmful level. Let's break this down:

Step 1: Understand Exponential Growth
Exponential growth can be expressed by the formula: , where is the final concentration, is the initial concentration, is the time in minutes, and is the doubling time (20 minutes here).

Step 2: Define Parameters
Initial concentration, .
Final harmful concentration, .
Doubling time, .

Step 3: Formulate the Equation

Step 4: Solve for
First, divide both sides by 5:
.
Take the logarithm on both sides:
.
Using the logarithmic identity, :
.
Solve for :
.

Step 5: Round to Nearest Integer
Rounding 219.09 gives us 219 minutes.

Step 1: Analysis of the graph:

1. [R] decreases rapidly - R is consumed quickly
2. [S] shows a peak then decreases - S is an intermediate that forms and then gets consumed
3. [Q] increases then levels off - Q accumulates as an intermediate
4. [P] increases steadily - P is the final product

Step 2: Key observations:

• S peaks early and stays low → S is consumed faster than it's produced → k₃ > k₁
• Q accumulates significantly → Q is produced faster than it's consumed → k₂ > k₄
• R depletes faster via k₂ than k₁ (since Q accumulates more than S) → k₂ > k₁

Answer: (A), (B), and (C) are correct.

The general rate expression for the reaction is:

$ \text{2. Effect of Doubling } [\text{H}_2\text{PO}_2^-] \text{H}_2\text{PO}_2^- [\text{H}_2\text{PO}_2^-]_{\text{new}} = 2 \times [\text{H}_2\text{PO}_2^-]_{\text{original}} \text{OH}^- [\text{OH}^-]_{\text{new}} = [\text{OH}^-]_{\text{original}} \text{Rate}_2 \text{3. Comparison of Rates} \text{Rate}_2 \text{Rate}_1 k [\text{H}_2\text{PO}_2^-] [\text{OH}^-]^2 \text{Conclusion} \text{H}_2\text{PO}_2^- 1 \text{H}_2\text{PO}_2^- \text{OH}^- $

For a zero-order reaction, the concentration of reactant P follows the equation:

Where:

• is the initial concentration of P,
• is the rate constant,
• is the time.

We are told that the concentration of P becomes half of its initial concentration in 30 minutes. Therefore, we can express this as:

Solving for k:

Now, to find the time when [P] becomes zero:

Substituting the value of k:

Solving for t:

Thus, the concentration of P becomes zero at 60 minutes.

The given function is .

To find the maximum, we first compute the derivative of the function:

Setting to find the critical points:

Since , we have:

Thus, the function attains a maximum at .

The molecular weight of PO can be calculated from the atomic weights of phosphorus (P) and oxygen (O). The molar mass of PO is:

Now, if 0.2 mol of P produces PO, the mass of PO produced is:

Thus, the mass of PO produced is 56.8 g

The reaction described is a nucleophilic substitution, likely following an SN2 mechanism, where the reaction rate is determined by the concentration of both the electrophile (in this case, the bromine compound) and the nucleophile (KCN).

In an SN₂ reaction, the rate law is typically of the form Rate = k[electrophile][nucleophile]. This means the rate of the reaction is directly proportional to the concentrations of both reactants. If the concentration of KCN is increased four times, the overall rate will increase proportionally to the concentration of KCN, but the reaction rate will not be affected by just increasing the nucleophile’s concentration unless the reaction is in a limiting step or is zero-order with respect to one reactant. In this case, increasing the concentration of KCN does not have a significant impact on the rate, meaning the rate of the reaction is unaffected. Hence, option (A) is the correct answer.

Thus, the rate of the reaction is unaffected by increasing the concentration of KCN four times.