Difficult
The mechanism of the reaction,$2NO_{(g)} + 2H_{2(g)} \to N_{2(g)} + 2H_2O_{(g)}$ is:
Step $1$: $2NO_{(g)} + H_{2(g)} \xrightarrow{\text{slow}} N_2 + H_2O_2$
Step $2$: $H_2O_2 + H_2 \xrightarrow{\text{fast}} 2H_2O$
Then the correct statement is:
Step $1$: $2NO_{(g)} + H_{2(g)} \xrightarrow{\text{slow}} N_2 + H_2O_2$
Step $2$: $H_2O_2 + H_2 \xrightarrow{\text{fast}} 2H_2O$
Then the correct statement is:
- ARate $= k [NO]^2 [H_2]$
- BRate $= k [H_2O_2] [H_2]$
- COn doubling the concentration of $H_2$,keeping the concentration of $NO$ constant,the rate will become double
- DIf the initial concentration of $H_2$ and $NO$ is $C_0$ and after time $t$ the concentration of $N_2$ is $x$,then Rate $= k (C_0 - 2x)^2$
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Rate law for a reaction between $A$ and $B$ is given by $R=k[A]^{n}[B]^{m}$. If concentration of $A$ is doubled and concentration of $B$ is halved from their initial value,the ratio of new rate of reaction to the initial rate of reaction $\left(\frac{r_2}{r_1}\right)$ is
For the reaction $A + B \rightarrow C$,we have the following data:
Initial concentration of $A$ $(M)$ Initial concentration of $B$ $(M)$ Initial Rate $(M \cdot s^{-1})$ $1$ $10$ $100$ $1$ $1$ $1$ $10$ $1$ $10$
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| $1$ | $10$ | $100$ |
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| $10$ | $1$ | $10$ |
The order of the reaction with respect to $A$ and $B$ are:
In the given reaction sequence,if $K_3 > K_2 > K_1$,then the rate-determining step is:
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