Question 12 Marks
Answer the following:
$(a)$ Give two examples of non-chemical process which obeys the first order kinetics.
$(b)$ The following results have been obtained during the kinetic studies for the reaction:
$P+2Q \rightarrow R + 2S$
Determine the rate law expression for the reaction.
$(a)$ Give two examples of non-chemical process which obeys the first order kinetics.
$(b)$ The following results have been obtained during the kinetic studies for the reaction:
$P+2Q \rightarrow R + 2S$
| Exp. | Initial P(mol/L) | Initial Q (mol/L) | Init. Rate of Formation of R(M min-1) |
| $1$ | $0.10$ | $0.10$ | $3.0 \times 10^{-4}$ |
| $2$ | $0.30$ | $0.30$ | $9.0 \times 10^{-4}$ |
| $3$ | $0.10$ | $0.30$ | $3.0 \times 10^{-4}$ |
| $4$ | $0.10$ | $0.40$ | $6.0 \times 10^{-4}$ |
Answer
View full question & answer→Answer the following:
$(i)\ i.$ Rate of growth of population, if there is no change of birth and death rates.
$ii.$ Radioactive decay in which the number of nuclei disintegrating is proportional to the number of nuclei present.
$(ii)$ Let the rate law expression be rate $= k[P] [Q]$ from the table we know that
Rate $1 = 3.0 \times 10^{-4} = k (0.10)^x (0.10)^y$
Rate $2 = 9.0 \times 10^{-4} = k(0.30)^x (0.30)^y$
$\text { Rate } 3=3.0 \times 10^{-4}= k (0.10)^{ x }(0.30)^{ y }$
$\frac{\text { Rate } 1}{\text { Rate } 3}=\left(\frac{1}{3}\right)^y \text { or } 1=\left(\frac{1}{3}\right)^y$
$\text { So } y =0$
$\frac{\text { Rate } 2}{\text { Rate } 3}=(3)^{ x } \text { or } 3=(3)^{ y }$
$\text { So } x =1$
$\text { Rate }= k [ P ]$
$(i)\ i.$ Rate of growth of population, if there is no change of birth and death rates.
$ii.$ Radioactive decay in which the number of nuclei disintegrating is proportional to the number of nuclei present.
$(ii)$ Let the rate law expression be rate $= k[P] [Q]$ from the table we know that
Rate $1 = 3.0 \times 10^{-4} = k (0.10)^x (0.10)^y$
Rate $2 = 9.0 \times 10^{-4} = k(0.30)^x (0.30)^y$
$\text { Rate } 3=3.0 \times 10^{-4}= k (0.10)^{ x }(0.30)^{ y }$
$\frac{\text { Rate } 1}{\text { Rate } 3}=\left(\frac{1}{3}\right)^y \text { or } 1=\left(\frac{1}{3}\right)^y$
$\text { So } y =0$
$\frac{\text { Rate } 2}{\text { Rate } 3}=(3)^{ x } \text { or } 3=(3)^{ y }$
$\text { So } x =1$
$\text { Rate }= k [ P ]$
