For the reaction,$2 N_2O_{5(g)} \rightarrow 4 NO_{2(g)} + O_{2(g)}$,$N_2O_5$ disappears at a rate of $x \ mol \ dm^{-3} \ s^{-1}$. Find the rate of formation of $O_2$?
- A$x \ mol \ dm^{-3} \ s^{-1}$
- B$2x \ mol \ dm^{-3} \ s^{-1}$
- C$\frac{x}{2} \ mol \ dm^{-3} \ s^{-1}$
- D$\frac{3x}{2} \ mol \ dm^{-3} \ s^{-1}$
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Consider the following reaction: $N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$. The rate of this reaction in terms of $N_2$ at $T \ K$ is $\frac{-d[N_2]}{dt} = 0.02 \ mol \ L^{-1} \ s^{-1}$. What is the value of $\frac{-d[H_2]}{dt}$ (in units of $mol \ L^{-1} \ s^{-1}$) at the same temperature?
Find the rate of the following reaction: $2 \ N_2O_{5(g)} \rightarrow 4 \ NO_{2(g)} + O_{2(g)}$ if the concentration of $NO_2$ increases to $5.2 \times 10^{-3} \ M$ in $100 \ s$.
For the reaction,$3 I^{-} + S_2 O_8^{2-} \rightarrow I_3^{-} + 2 SO_4^{2-}$,at a particular time $t$,$\frac{d[SO_4^{2-}]}{dt}$ is $2.2 \times 10^{-2} \ mol \ dm^{-3} \ s^{-1}$. What is the value of $-\frac{d[I^{-}]}{dt}$?
Explain: What is instantaneous rate? How is it determined?
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View Solution$R \longrightarrow P$ is a first order reaction. The concentration of $R$ changed from $0.04$ to $0.03 \ mol \ L^{-1}$ in $40 \ min$. What is the average velocity of the reaction in $mol \ L^{-1} \ s^{-1}$?
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