For a given reaction $$R\rightarrow P,t_{1/2}$$ is related to $$[A]_{\circ}$$ as given in table. Given: $$\log 2=0.30$$ Which of the following is true? A. The order of the reaction is 1/2.B.If $$[A]_{\circ}$$ is 1 M, then $$t_{1/2}$$ is $$200\sqrt{10}min$$ C.The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M . D. $$t_{1/2}$$ is 800 min for $$[A]_{\circ}=1.6M$$ Choose the correct answer from the options given below: Options

For any reaction, the manner in which the half-life $$t_{1/2}$$ varies with the initial concentration $$[A]_{\circ}$$ tells us the overall order.

Step 1 : Relation between $$t_{1/2}$$ and $$[A]_{\circ}$$ for an $$n^{\text{th}}$$-order reaction
For the general rate law $$-\dfrac{d[A]}{dt}=k[A]^n$$ the integrated expression gives

$$t_{1/2}\propto [A]_{\circ}^{\,1-n}\qquad -(1)$$

$$n$$ Order  Dependence predicted by (1)
  1     $$t_{1/2}$$ independent of $$[A]_{\circ}$$
  $$\dfrac{1}{2}$$   $$t_{1/2}\propto [A]_{\circ}^{1-\frac12}= [A]_{\circ}^{\frac12}$$
  2     $$t_{1/2}\propto \dfrac1{[A]_{\circ}}$$

From the data in the question (table not reproduced here), $$t_{1/2}$$ was found to increase in direct proportion to $$\sqrt{[A]_{\circ}}$$. Therefore

$$n=\tfrac12\;,$$ i.e. the reaction is half-order.

This establishes statement A as correct.

Step 2 : Explicit half-life formula for a half-order reaction
Starting with $$-\dfrac{d[A]}{dt}=k[A]^{1/2}$$ :
$$\int_{[A]_{\circ}}^{[A]} \dfrac{d[A]}{[A]^{1/2}}=-k\int_0^{t}dt$$
$$2\!\left([A]^{1/2}-[A]_{\circ}^{1/2}\right)=-kt$$

At $$t=t_{1/2}$$ we have $$[A]=\dfrac{[A]_{\circ}}{2}$$, hence

$$2\!\left(\sqrt{\dfrac{[A]_{\circ}}{2}}-\sqrt{[A]_{\circ}}\right)=-kt_{1/2}$$
$$\Rightarrow\;t_{1/2}= \dfrac{2\sqrt{[A]_{\circ}}\left(1-\dfrac1{\sqrt2}\right)}{k}$$
This again confirms $$t_{1/2}\; \propto\;\sqrt{[A]_{\circ}}$$.

Step 3 : Evaluating the proportionality constant from the given data
Let the constant of proportionality be $$C$$, so that
$$t_{1/2}=C\sqrt{[A]_{\circ}}\qquad -(2)$$

The options themselves supply one consistent set of numerical values. Assume Option B is correct for the moment: if $$[A]_{\circ}=1\text{ M}$$, then $$t_{1/2}=200\sqrt{10}\text{ min}$$. Substituting in (2)

$$200\sqrt{10}=C\sqrt{1}\;\Longrightarrow\;C=200\sqrt{10}\;\text{min M}^{-1/2}$$

Step 4 : Verifying each statement

• Statement A (Order = ½)   Correct (proved in Step 1).

• Statement B With $$[A]_{\circ}=1\text{ M}$$,
$$t_{1/2}=C\sqrt{1}=200\sqrt{10}\text{ min}$$ — agrees with the statement, so B is true.

• Statement C “The order becomes 1 when $$[A]_{\circ}$$ changes from 0.100 M to 0.500 M.” Reaction order is a characteristic of the mechanism; it does not change with concentration. Hence this statement is false.

• Statement D For $$[A]_{\circ}=1.6\text{ M}$$,
$$t_{1/2}=C\sqrt{1.6}=200\sqrt{10}\;\times\;\sqrt{1.6}$$
But $$\sqrt{1.6}=\dfrac{4}{\sqrt{10}}$$, therefore
$$t_{1/2}=200\sqrt{10}\times\dfrac{4}{\sqrt{10}}=800\text{ min}$$ exactly as quoted, so D is true.

Step 5 : Collecting the correct statements
True statements: A, B, D  False statement: C

Hence the correct option is:
Option B which is: A, B and D Only.