For the reaction $2N_2O_{5(g)} \to 4NO_{2(g)} + O_{2(g)}$,if the concentration of $NO_2$ increases by $5.2 \times 10^{-3} \ M$ in $100 \ s$,then the rate of reaction will be:
- A$1.3 \times 10^{-5} \ M \ s^{-1}$
- B$5 \times 10^{-4} \ M \ s^{-1}$
- C$7.6 \times 10^{-4} \ M \ s^{-1}$
- D$2 \times 10^{-3} \ M \ s^{-1}$
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Similar Questions
The instantaneous rate of disappearance of $MnO_4^-$ ion in the following reaction is $4.56 \times 10^{-3} \ M s^{-1}$.
$2MnO_4^- + 10I^{-} + 16H^{+} \to 2Mn^{2+} + 5I_2 + 8H_2O$
The rate of appearance of $I_2$ is $...... \times 10^{-2} \ M s^{-1}$.
$2MnO_4^- + 10I^{-} + 16H^{+} \to 2Mn^{2+} + 5I_2 + 8H_2O$
The rate of appearance of $I_2$ is $...... \times 10^{-2} \ M s^{-1}$.
DifficultJEE MAIN 2013
View SolutionFor the reaction $3 A + 2 B \rightarrow C$,which statement is correct $:-$
Medium
View SolutionIn a reaction $R \to P$,the concentration $[M]$ obtained at different times $(t)$ are shown in the following table. Calculate the average rate $r_{av}$ of the reaction.
Time $t$ $(s)$ $0$ $5$ $10$ $20$ $30$ Concentration $(mol \ L^{-1})$ $160 \times 10^{-3}$ $80 \times 10^{-3}$ $40 \times 10^{-3}$ $10 \times 10^{-3}$ $3.5 \times 10^{-3}$
| Time $t$ $(s)$ | $0$ | $5$ | $10$ | $20$ | $30$ |
| Concentration $(mol \ L^{-1})$ | $160 \times 10^{-3}$ | $80 \times 10^{-3}$ | $40 \times 10^{-3}$ | $10 \times 10^{-3}$ | $3.5 \times 10^{-3}$ |
Medium
View SolutionNitric oxide reacts with $H_2$ according to the reaction: $2 NO_{(g)} + 2 H_{2(g)} \rightarrow N_{2(g)} + 2 H_2 O_{(g)}$. Identify the correct relation for the rate of disappearance of reactants and the rate of appearance of products.
Find the average rate of formation of $NO_{2_{(g)}}$ in the following reaction:
$2 N_2O_{5_{(g)}} \rightarrow 4 NO_{2_{(g)}} + O_{2_{(g)}}$
Given that $\left[ -\frac{\Delta [N_2O_5]}{\Delta t} \right] = x \ mol \ dm^{-3} \ s^{-1}$
$2 N_2O_{5_{(g)}} \rightarrow 4 NO_{2_{(g)}} + O_{2_{(g)}}$
Given that $\left[ -\frac{\Delta [N_2O_5]}{\Delta t} \right] = x \ mol \ dm^{-3} \ s^{-1}$
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