Question

A solid spherical shape is assumed to react only on its surface However the amount of reactant contained in it is determined by its volume. Consider the following rate law for a solid shrinking due to reaction:
 $ \dfrac{{ - dV}}{{\;\;dt}}\; = \;K\;\left( {S = surface\;area} \right) $
The order of above reaction will be:
(A) $ 0 $
(B) $ 1 $
(C) $ 1/3 $
(D) $ 3/2 $

Answer 1

Hint: We are calculating rate of the reaction we should know about rate of a reaction. Rate of a reaction is defined as the speed at which the reactants in a given reaction are converted into products. It gives us the idea about the time taken by the reaction to be completed. In this question the concentration of reactant depends on the Volume of the solid sphere.

Complete answer:
Rate of reaction is given by
 $ {\text{r}}\; = \;{\text{k}}{\left[ {\text{X}} \right]^{\text{a}}} $
where r is the rate of the reaction, k is the rate constant, X is the concentration of reactant X and a is the order of the reaction.
For our question initial rate of reaction is given by
 $ {{\text{r}}_1}\; = \;{\text{k}}{\left[ {{{\text{X}}_1}} \right]^{\text{n}}} $ ………. $ \left( 1 \right) $
let’s assume the concentration of the reactant to be X, n is the required order of the reaction
If we double the concentration of the reactant than rate of reaction will be doubled as the concentration of product will be doubled, writing the rate of reaction for doubled concentration of reactant
 $ {{\text{r}}_2}\; = \;{\text{k}}{\left[ {{{\text{X}}_2}} \right]^{\text{n}}} $
 $ 2{{\text{r}}_1}\; = \;{\text{k}}{\left[ {{{\text{X}}_2}} \right]^{\text{n}}} $
Since the concentration of reactant is depending on volume the concentration of reactant becomes eight times of initial concentration
 $ 2{{\text{r}}_1}\; = \;{\text{k}}{\left[ {8{{\text{X}}_1}} \right]^{\text{n}}} $ ………….. $ \left( 2 \right) $
Dividing equation $ \left( 2 \right) $ by equation $ \left( 1 \right) $ we get
 $ \dfrac{{2{{\text{r}}_1}}}{{{{\text{r}}_1}}}\; = \;\dfrac{{{\text{k}}{{\left[ {8{{\text{X}}_1}} \right]}^{{\text{n}}\;}}\;\;}}{{{\text{k}}{{\left[ {{{\text{X}}_1}} \right]}^{\text{n}}}}} $
simplifying the equation, we get
 $ 2\; = \;{8^{\text{n}}}\; $
 $ {\text{n}} = \dfrac{1}{3}\;\; $
The order of the given reaction is $ 1/3 $ Hence option (C) is correct.

Note:
Volume of a sphere is given by $ {\text{V}}\; = \;\dfrac{4}{3}\pi {r^3}\; $ , the reaction is taking place on the surface and concentration of reactant depends on the volume, volume becomes eight times the initial volume when we double the radius that is why concentration of reactant become eight times the initial concentration.