For the reaction,$2 NO_{2(g)} \rightarrow 2 NO_{(g)} + O_{2(g)}$,if $\frac{-d[NO_2]}{dt}$ is $1.3 \times 10^{-5} \ mol \ L^{-1} \ sec^{-1}$,what is the rate of formation of $O_2$?
- A$1.3 \times 10^{-6} \ mol \ L^{-1} \ sec^{-1}$
- B$3.25 \times 10^{-6} \ mol \ L^{-1} \ sec^{-1}$
- C$1.62 \times 10^{-6} \ mol \ L^{-1} \ sec^{-1}$
- D$6.5 \times 10^{-6} \ mol \ L^{-1} \ sec^{-1}$
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Write the rate equation for the following reactions:
$1.$ $5 Br^{-}_{(aq)} + BrO^{-}_{3_{(aq)}} + 6 H^{+}_{(aq)} \rightarrow 3 Br_{2_{(aq)}} + 3 H_2O_{(l)}$
$2.$ $R \rightarrow P$
$3.$ $C_4H_9Cl + H_2O \rightarrow C_4H_9OH + HCl$
$1.$ $5 Br^{-}_{(aq)} + BrO^{-}_{3_{(aq)}} + 6 H^{+}_{(aq)} \rightarrow 3 Br_{2_{(aq)}} + 3 H_2O_{(l)}$
$2.$ $R \rightarrow P$
$3.$ $C_4H_9Cl + H_2O \rightarrow C_4H_9OH + HCl$
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View SolutionIn the following reaction: $xA \to yB$,$\log \left[ -\frac{d[A]}{dt} \right] = \log \left[ \frac{d[B]}{dt} \right] + \log 2$,where the negative sign indicates the rate of disappearance of the reactant. Thus,$x : y$ is:
In the hydrogenation of ethene at $298 \, K$,the number of moles of hydrogen decreases from $2.2$ to $1.4$ in $50 \, min$. Calculate the rate of reaction in units of $mol \, s^{-1}$.
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View SolutionAmmonia and oxygen react at high temperature as $4 NH_{3(g)} + 5 O_{2(g)} \longrightarrow 4 NO_{(g)} + 6 H_2 O_{(g)}$. If the rate of formation of $NO_{(g)}$ is $3.6 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$,then the rate of disappearance of ammonia is:
If $3A \to 2B$,then the rate of reaction of $+\frac{d[B]}{dt}$ is equal to
MediumAIPMT 2002
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