The concentration of a reactant changes from $0.1 \, mol \, L^{-1}$ to $0.095 \, mol \, L^{-1}$ in $10 \, s$. Calculate the rate of reaction.
- A$5.0 \times 10^{-4} \, mol \, L^{-1} \, s^{-1}$
- B$5.0 \times 10^{-3} \, mol \, L^{-1} \, s^{-1}$
- C$5.0 \times 10^{-5} \, mol \, L^{-1} \, s^{-1}$
- D$1.0 \times 10^{-4} \, mol \, L^{-1} \, s^{-1}$
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The rate of reaction is expressed as
$+\frac{1}{2}\frac{d[C]}{dt} = -\frac{1}{3}\frac{d[D]}{dt} = +\frac{1}{4}\frac{d[A]}{dt} = -\frac{d[B]}{dt}$
The reaction is:
$+\frac{1}{2}\frac{d[C]}{dt} = -\frac{1}{3}\frac{d[D]}{dt} = +\frac{1}{4}\frac{d[A]}{dt} = -\frac{d[B]}{dt}$
The reaction is:
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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:
For the reaction $2 N_2O_{5(g)} \rightarrow 4 NO_{2(g)} + O_{2(g)}$,the initial concentration of $N_2O_5$ is $2.0 \ mol \ L^{-1}$ and after $300 \ min$,it is reduced to $1.4 \ mol \ L^{-1}$. The rate of production of $NO_2$ (in $mol \ L^{-1} \ min^{-1}$) is
MediumKCET 2025
View SolutionThe instantaneous rate for the reaction $2 A + B \rightarrow C + 3 D$ is given by:
In a reaction,$n_1A + n_2B \to m_1C + m_2D$,$5 \ M$ of reactant $A$ are allowed to react with $3 \ M$ of reactant $B$. After $5 \ sec$,the concentration of $A$ was found to be $4 \ M$. What are the rate of decomposition of $A$ and the rate of formation of $D$,respectively?
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