Medium
For the decomposition reaction $2NH_3(g) \rightarrow N_2(g) + 3H_2(g)$,if the rate constant $k = 2.6 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$,calculate the rate of formation of $N_2$ and $H_2$.
The rate of reaction is given by: $\text{Rate} = -\frac{1}{2} \frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt}$.
Since the reaction is zero order,$\text{Rate} = k = 2.6 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
For $N_2$: $\frac{d[N_2]}{dt} = \text{Rate} = 2.6 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
For $H_2$: $\frac{1}{3} \frac{d[H_2]}{dt} = \text{Rate}$ $\Rightarrow \frac{d[H_2]}{dt} = 3 \times \text{Rate} = 3 \times (2.6 \times 10^{-4}) = 7.8 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
Since the reaction is zero order,$\text{Rate} = k = 2.6 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
For $N_2$: $\frac{d[N_2]}{dt} = \text{Rate} = 2.6 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
For $H_2$: $\frac{1}{3} \frac{d[H_2]}{dt} = \text{Rate}$ $\Rightarrow \frac{d[H_2]}{dt} = 3 \times \text{Rate} = 3 \times (2.6 \times 10^{-4}) = 7.8 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$.
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Similar Questions
The reaction rate between two substances $A$ and $B$ is expressed as: $\text{rate} = k[A]^n[B]^m$. If the concentration of $A$ is doubled and the concentration of $B$ is halved,the ratio of the new rate to the initial rate will be:
MediumAIEEE 2012
View SolutionThe initial rates of decrease of $I_2$ in acetone-iodine reaction catalysed by $H^{+}$ are given in the table.
Experiment Initial $[I_2]$ $(mol \ L^{-1})$ Initial $[H^{+}]$ $(mol \ L^{-1})$ Initial $[CH_3COCH_3]$ $(mol \ L^{-1})$ Initial rate $(mol \ L^{-1} \ s^{-1})$ $1$ $0.01$ $0.1$ $0.1$ $0.096$ $2$ $0.01$ $0.2$ $0.1$ $0.192$ $3$ $0.02$ $0.2$ $0.1$ $0.192$ $4$ $0.01$ $0.2$ $0.2$ $0.384$
The order with respect to $I_2, H^{+}$,acetone and total order of the reaction respectively are:
| Experiment | Initial $[I_2]$ $(mol \ L^{-1})$ | Initial $[H^{+}]$ $(mol \ L^{-1})$ | Initial $[CH_3COCH_3]$ $(mol \ L^{-1})$ | Initial rate $(mol \ L^{-1} \ s^{-1})$ |
| $1$ | $0.01$ | $0.1$ | $0.1$ | $0.096$ |
| $2$ | $0.01$ | $0.2$ | $0.1$ | $0.192$ |
| $3$ | $0.02$ | $0.2$ | $0.1$ | $0.192$ |
| $4$ | $0.01$ | $0.2$ | $0.2$ | $0.384$ |
The order with respect to $I_2, H^{+}$,acetone and total order of the reaction respectively are:
For the reaction $A + B \to \text{products}$,what will be the order of reaction with respect to $A$ and $B$?
$Exp.$ $[A] \ (mol \ L^{-1})$ $[B] \ (mol \ L^{-1})$ Initial rate $(mol \ L^{-1} \ s^{-1})$ $1$ $2.5 \times 10^{-4}$ $3 \times 10^{-5}$ $5 \times 10^{-4}$ $2$ $5 \times 10^{-4}$ $6 \times 10^{-5}$ $4 \times 10^{-3}$ $3$ $1 \times 10^{-3}$ $6 \times 10^{-5}$ $1.6 \times 10^{-2}$
| $Exp.$ | $[A] \ (mol \ L^{-1})$ | $[B] \ (mol \ L^{-1})$ | Initial rate $(mol \ L^{-1} \ s^{-1})$ |
| $1$ | $2.5 \times 10^{-4}$ | $3 \times 10^{-5}$ | $5 \times 10^{-4}$ |
| $2$ | $5 \times 10^{-4}$ | $6 \times 10^{-5}$ | $4 \times 10^{-3}$ |
| $3$ | $1 \times 10^{-3}$ | $6 \times 10^{-5}$ | $1.6 \times 10^{-2}$ |
Match List-$I$ with List-$II$:
List-$I$ (Order of Reaction) List-$II$ (Unit of rate constant) $A$. Zero order $I$. $mol^{-1} L s^{-1}$ $B$. First order $II$. $mol^{-2} L^{2} s^{-1}$ $C$. Second order $III$. $s^{-1}$ $D$. Third order $IV$. $mol L^{-1} s^{-1}$
| List-$I$ (Order of Reaction) | List-$II$ (Unit of rate constant) |
| $A$. Zero order | $I$. $mol^{-1} L s^{-1}$ |
| $B$. First order | $II$. $mol^{-2} L^{2} s^{-1}$ |
| $C$. Second order | $III$. $s^{-1}$ |
| $D$. Third order | $IV$. $mol L^{-1} s^{-1}$ |
MediumNEET 2026
View Solution$A$ reaction involving two different reactants:
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