Difficult
$A \rightarrow$ products ($1^{st}$ order reaction). Three sets of experiment were performed for a reaction under similar experimental conditions. Run $1 \Rightarrow 100 \ mL$ of $10 \ M$ solution of reactant $A$. Run $2 \Rightarrow 200 \ mL$ of $10 \ M$ solution of reactant $A$. Run $3 \Rightarrow 100 \ mL$ of $10 \ M$ solution of reactant $A + 100 \ mL$ of $H_2O$ added. The correct variation of rate of reaction is:
- ARun $1 = $ Run $2 = $ Run $3$
- BRun $3 < $ Run $1 = $ Run $2$
- CRun $3 < $ Run $1 < $ Run $2$
- DRun $1 < $ Run $2 < $ Run $3$
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Similar Questions
Write the differential rate expression for the following reactions and determine their order of reaction:
$1) \ 2 N_2O_5(g) \rightarrow 4 NO_2(g) + O_2(g)$
$2) \ C_4H_9Cl(aq) + OH^-(aq) \rightarrow C_4H_9OH(aq) + Cl^-(aq)$
$1) \ 2 N_2O_5(g) \rightarrow 4 NO_2(g) + O_2(g)$
$2) \ C_4H_9Cl(aq) + OH^-(aq) \rightarrow C_4H_9OH(aq) + Cl^-(aq)$
Medium
View SolutionConsider the following reaction $A + B \rightarrow C$.
The time taken for $A$ to become $1/4$ of its initial concentration is twice the time taken to become $1/2$ of the same. Also,when the change of concentration of $B$ is plotted against time,the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis. The overall order of the reaction is . . . . .
The time taken for $A$ to become $1/4$ of its initial concentration is twice the time taken to become $1/2$ of the same. Also,when the change of concentration of $B$ is plotted against time,the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis. The overall order of the reaction is . . . . .
The following data was obtained for the chemical reaction given below at $975 \ K$: $2 NO_{(g)} + 2 H_{2(g)} \rightarrow N_{2(g)} + 2 H_{2}O_{(g)}$
Experiment $[NO] \ (mol \ L^{-1})$ $[H_{2}] \ (mol \ L^{-1})$ Rate $(mol \ L^{-1} \ s^{-1})$ $1$ $8 \times 10^{-5}$ $8 \times 10^{-5}$ $7 \times 10^{-9}$ $2$ $24 \times 10^{-5}$ $8 \times 10^{-5}$ $2.1 \times 10^{-8}$ $3$ $24 \times 10^{-5}$ $32 \times 10^{-5}$ $8.4 \times 10^{-8}$
The order of the reaction with respect to $NO$ is ..... .
| Experiment | $[NO] \ (mol \ L^{-1})$ | $[H_{2}] \ (mol \ L^{-1})$ | Rate $(mol \ L^{-1} \ s^{-1})$ |
| $1$ | $8 \times 10^{-5}$ | $8 \times 10^{-5}$ | $7 \times 10^{-9}$ |
| $2$ | $24 \times 10^{-5}$ | $8 \times 10^{-5}$ | $2.1 \times 10^{-8}$ |
| $3$ | $24 \times 10^{-5}$ | $32 \times 10^{-5}$ | $8.4 \times 10^{-8}$ |
$2 \ NO + 2 \ H_2 \rightarrow N_2 + 2 \ H_2O$
The above reaction has been studied at $800^{\circ} C$. The related data are given in the table below.
Reaction serial number Initial pressure of $H_2$ / $kPa$ Initial Pressure of $NO$ / $kPa$ Initial rate $(-\frac{dp}{dt}) / (kPa \ s^{-1})$ $1$ $65.6$ $40.0$ $0.135$ $2$ $65.6$ $20.1$ $0.033$ $3$ $38.6$ $65.6$ $0.214$ $4$ $19.2$ $65.6$ $0.106$
The order of the reaction with respect to $NO$ is $...........$
The above reaction has been studied at $800^{\circ} C$. The related data are given in the table below.
| Reaction serial number | Initial pressure of $H_2$ / $kPa$ | Initial Pressure of $NO$ / $kPa$ | Initial rate $(-\frac{dp}{dt}) / (kPa \ s^{-1})$ |
| $1$ | $65.6$ | $40.0$ | $0.135$ |
| $2$ | $65.6$ | $20.1$ | $0.033$ |
| $3$ | $38.6$ | $65.6$ | $0.214$ |
| $4$ | $19.2$ | $65.6$ | $0.106$ |
The order of the reaction with respect to $NO$ is $...........$
The rate law expression for the reaction $aA + bB \to P$ is $\text{rate} = K [A]^p [B]^q$. The order of the reaction is
Medium
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