(a) Half-life period of a reaction increases with increase in initial concentration. Predict the order of the reaction. (b) Decomposition of a compound follows first-order kinetics. If it takes 15 minutes for 20% of original substance to decay, calculate -- (i) rate constant; (ii

Question

(a) Half-life period of a reaction increases with increase in initial concentration. Predict the order of the reaction. (b) Decomposition of a compound follows first-order kinetics. If it takes 15 minutes for 20% of original substance to decay, calculate -- (i) rate constant; (ii) the time at which 10% of the reactant remains undecayed.

**OR**

(c) Define activation energy. (d) The rate of a particular reaction doubles when temperature changes from 27 C to 37 C. Calculate the value of activation energy. ($R = 8.314\ 	ext{J K}^{-1}	ext{mol}^{-1}$)

Accepted Answer

>

> [!TLDR]
> (a)/(b) use the first-order half-life-vs-concentration relation and the integrated first-order rate law. (OR) (c)/(d) use the definition of activation energy and the two-point Arrhenius equation.
>
> **(a) Order from half-life behaviour**
>
> In general, $t_{1/2} \propto [A]_0^{\,1-n}$ for a reaction of order $n$. If $t_{1/2}$ **increases** as the initial concentration increases, the exponent $(1-n)$ must be positive, i.e. $n < 1$. This is exactly the behaviour of a **zero-order** reaction, where $t_{1/2} = \dfrac{[A]_0}{2k}$ — directly proportional to $[A]_0$. (Contrast: first order gives a *constant* $t_{1/2}=0.693/k$ independent of concentration; second order gives $t_{1/2}$ that *decreases* with concentration.)
>
> So the reaction is **zero order**.
>
> **(b) First-order decomposition, 20% decayed in 15 minutes**
>
> (i) After 15 min, 80% of the original substance remains: $[A]/[A]_0 = 0.80$.
>
> $$k = \frac{2.303}{t}\log\frac{[A]_0}{[A]} = \frac{2.303}{15}\log\left(\frac{1}{0.80}ight) = \frac{2.303}{15}	imes 0.0969$$
>
> $$k = 0.1535 	imes 0.0969 \approx 1.49	imes10^{-2}\ 	ext{min}^{-1}$$
>
> (ii) Time for only 10% to remain undecayed, i.e. $[A]/[A]_0 = 0.10$:
>
> $$t = \frac{2.303}{k}\log\frac{[A]_0}{[A]} = \frac{2.303}{1.49	imes10^{-2}}\log(10) = \frac{2.303}{0.0149}	imes 1$$
>
> $$t \approx 154.8\ 	ext{minutes} \ (\approx 2.58\ 	ext{hours})$$
>
> **OR — (c) Activation energy**
>
> Activation energy ($E_a$) is the **minimum extra energy** (above the average energy of the reactant molecules) that colliding molecules must possess for a collision to be effective and lead to product formation — equivalently, the energy required to reach the top of the energy barrier (the transition state / activated complex) that separates reactants from products.
>
> **(d) Activation energy from the Arrhenius equation**
> …

Q.(a) Half-life period of a reaction increases with increase in initial concentration. Predict the order of the reaction.

Concept understanding — Integrated Rate Law First Order

Integrated Rate Law for a First-Order Reaction

The Intuition

The Derivation (Short)

The Linear Plot — The Key Insight

What This Means Practically

Why the Arrhenius plot idea works here

Step-by-step derivation

Why the Arrhenius Plot Works

Step-by-Step Derivation

Q.(a) Half-life period of a reaction increases with increase in initial concentration. Predict the order of the reaction.

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