Medium
Consider the following reactions:
$A$ $\rightarrow P_1; B$ $\rightarrow P_2; C$ $\rightarrow P_3; D$ $\rightarrow P_4$
The orders of the above reactions are $a, b, c,$ and $d,$ respectively. The following graph is obtained when $\log[\text{rate}]$ vs. $\log[\text{conc.}]$ are plotted:
Among the following,the correct sequence for the order of the reactions is:
$A$ $\rightarrow P_1; B$ $\rightarrow P_2; C$ $\rightarrow P_3; D$ $\rightarrow P_4$
The orders of the above reactions are $a, b, c,$ and $d,$ respectively. The following graph is obtained when $\log[\text{rate}]$ vs. $\log[\text{conc.}]$ are plotted:
Among the following,the correct sequence for the order of the reactions is:
- A$a > b > c > d$
- B$c > a > b > d$
- C$d > b > a > c$
- D$d > a > b > c$
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Difficult
View Solution$2 \ NO_{(g)} + Cl_{2(g)} \rightleftharpoons 2 \ NOCl_{(g)}$
This reaction was studied at $-10^{\circ} C$ and the following data was obtained:
$Run$ $[NO]_{0}$ $[Cl_{2}]_{0}$ $r_{0}$ $1$ $0.10$ $0.10$ $0.18$ $2$ $0.10$ $0.20$ $0.35$ $3$ $0.20$ $0.20$ $1.40$
$[NO]_{0}$ and $[Cl_{2}]_{0}$ are the initial concentrations and $r_{0}$ is the initial reaction rate.
The overall order of the reaction is ..........
(Round off to the Nearest Integer).
This reaction was studied at $-10^{\circ} C$ and the following data was obtained:
| $Run$ | $[NO]_{0}$ | $[Cl_{2}]_{0}$ | $r_{0}$ |
| $1$ | $0.10$ | $0.10$ | $0.18$ |
| $2$ | $0.10$ | $0.20$ | $0.35$ |
| $3$ | $0.20$ | $0.20$ | $1.40$ |
$[NO]_{0}$ and $[Cl_{2}]_{0}$ are the initial concentrations and $r_{0}$ is the initial reaction rate.
The overall order of the reaction is ..........
(Round off to the Nearest Integer).
The following results have been obtained during the kinetic studies of the reaction: $2 \ NO + 2 \ H_2 \longrightarrow N_2 + 2 \ H_2O$
Expt $\frac{-d[NO]}{dt} \ (mol \ L^{-1} \ s^{-1})$ $[NO] \ (mol \ L^{-1})$ $[H_2] \ (mol \ L^{-1})$ $1$ $4.8 \times 10^{-5}$ $1 \times 10^{-2}$ $1 \times 10^{-3}$ $2$ $43.2 \times 10^{-5}$ $3 \times 10^{-2}$ $1 \times 10^{-3}$ $3$ $86.4 \times 10^{-5}$ $3 \times 10^{-2}$ $2 \times 10^{-3}$
| Expt | $\frac{-d[NO]}{dt} \ (mol \ L^{-1} \ s^{-1})$ | $[NO] \ (mol \ L^{-1})$ | $[H_2] \ (mol \ L^{-1})$ |
| $1$ | $4.8 \times 10^{-5}$ | $1 \times 10^{-2}$ | $1 \times 10^{-3}$ |
| $2$ | $43.2 \times 10^{-5}$ | $3 \times 10^{-2}$ | $1 \times 10^{-3}$ |
| $3$ | $86.4 \times 10^{-5}$ | $3 \times 10^{-2}$ | $2 \times 10^{-3}$ |
For the reaction $NH_4^+ + NO_2^- \to N_2 + 2H_2O$,the experimental data is given below. Determine the rate law for the reaction.
$1.$ $[NH_4^+] = 0.24 \, M, [NO_2^-] = 0.10 \, M, \text{Rate} = 7.2 \times 10^{-6} \, M/s$ $2.$ $[NH_4^+] = 0.12 \, M, [NO_2^-] = 0.10 \, M, \text{Rate} = 3.6 \times 10^{-6} \, M/s$ $3.$ $[NH_4^+] = 0.12 \, M, [NO_2^-] = 0.15 \, M, \text{Rate} = 5.4 \times 10^{-6} \, M/s$
| $1.$ $[NH_4^+] = 0.24 \, M, [NO_2^-] = 0.10 \, M, \text{Rate} = 7.2 \times 10^{-6} \, M/s$ |
| $2.$ $[NH_4^+] = 0.12 \, M, [NO_2^-] = 0.10 \, M, \text{Rate} = 3.6 \times 10^{-6} \, M/s$ |
| $3.$ $[NH_4^+] = 0.12 \, M, [NO_2^-] = 0.15 \, M, \text{Rate} = 5.4 \times 10^{-6} \, M/s$ |
Medium
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