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Question 14 Marks
(a) Define molecularity of the reaction-
(b) Explain the effect of the presence of a catalyst on the rate of reaction.
(c) In a first order reaction at 300 k the initial quantity of the reactant was $1.0 X ^{-2} molL ^{-1}$ which reduced to $0.5 \times 10^{-2} molL ^{-1}$ in 30 minutes Calculate the rate constant of the reaction at 300 k .
Answer
(a) Definition of Molecularity
Molecularity of a reaction is defined as the total number of atoms, ions, or molecules of reactants that must collide simultaneously in an elementary (single-step) reaction to result in a chemical change. It is always a theoretical value, a positive whole number, and cannot be zero or fractional.
(b) Effect of Catalyst on Rate of Reaction
A catalyst increases the rate of a chemical reaction without undergoing any permanent chemical change itself. It functions by providing an alternative reaction pathway that has a lower activation energy ( $E_a$ ) compared to the uncatalyzed reaction.
By lowering the activation energy barrier, a larger fraction of reactant molecules possess sufficient energy to cross the threshold and form products, thereby significantly increasing the reaction velocity.
(c) Numerical Calculation
Given:
- Temperature ( $T$ ): 300 K (Constant)
- Initial Concentration $\left([A]_0\right): 1.0 \times 10^{-2} mol L ^{-1}$
- Final Concentration $([A]): 0.5 \times 10^{-2} mol L ^{-1}$
- Time $(t)$ : 30 minutes
Observation: Notice that the final concentration is exactly half of the initial concentration ( 0.5 is half of 1.0$)$. This means the given time $(30 min)$ is the half-life $\left(t_{1 / 2}\right)$ of the reaction.
Formula for First Order Rate Constant:
$k=\frac{0.693}{t_{1 / 2}}$
Calculation:
$\begin{array}{c}k=\frac{0.693}{30 min} \\ k=0.0231 min^{-1}\end{array}$
In Scientific Notation:
$k=2.31 \times 10^{-2} min^{-1}$
The rate constant of the reaction at 300 K is $2.31 \times 10^{-2} min^{-1}$.
Would you like me to show you how to convert this rate constant into seconds $\left(s^{-1}\right)$ or explain the difference between Order and Molecularity?
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Question 24 Marks
(a) In a chemical reaction rate constant almost doubles with the increase in temperature of $10^{\circ} C$ Explain with the help of a labelled distribution graph.
(b) The rate constant for a first order reaction is $60 Sec ^{-1}$ How long will it take for a Substance to become one sixteenth of its concentration.
Answer
(a)self
(b)Numerical Solution
Given:
- Rate constant $(k)=60 s^{-1}$
- Final concentration $([A])=\frac{1}{16}$ of initial concentration $\left([A]_0\right)$
Formula for First Order Reaction: The time $(t)$ required for a first-order reaction is calculated using the integrated rate equation:
$t=\frac{2.303}{k} \log \frac{[A]_0}{[A]}$
Step-by-Step Calculation:
1. Substitute the concentrations: Since $[A]=\frac{[A]_0}{16}$, the ratio $\frac{[A]_0}{A]}$ becomes 16 .
2. Apply the values in the formula:
$t=\frac{2.303}{60} \log (16)$
3. Calculate the log value: $\log (16)=\log \left(2^4\right)=4 \times \log (2)$ Since $\log (2) \approx 0.3010$, then $\log (16) \approx 1.2041$.
4. Final Computation:
$\begin{array}{c}t=\frac{2.303 \times 1.2041}{60} \\ t=\frac{2.773}{60}\end{array}$
$t \approx 0.0462$ seconds
It will take approximately $4.62 \times 10^{-2}$ seconds for the substance to reduce to one-sixteenth of its initial concentration.
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