The rate equation for a first-$order$ reaction is given by $R = R_0 e^{-kt}$. $A$ straight line with positive slope is obtained by plotting which of the following? ($R_0 =$ Initial concentration of the reactant,$R =$ concentration of the reactant at time $t$)
- A$\log(R_0/R)$ vs time
- B$R$ vs time
- C$\log R$ vs time
- D$\log(R/R_0)$ vs time
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For a first order reaction,the intercept of the graph between $\log \left(\frac{[A]_0}{[A]_t}\right)$ ($Y$-axis) and time ($X$-axis) is equal to
$A$ first-order reaction takes $69.3 \ min$ to complete $50\%$ of the reaction. How much time (in $min$) will it take to complete $80\%$ of the reaction?
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View SolutionIf the half-life of a first-order reaction is $6.93$ minutes, then the time required for the completion of $99\%$ of the reaction will be . . . . . . minutes. (Given: $\log 2 = 0.3010$)
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View SolutionIf $60\%$ of a first order reaction was completed in $60 \ min$,$50\%$ of the same reaction would be completed in approximately $.......... \ min.$
$(\log \, 4 = 0.60, \, \log \, 5 = 0.69)$
$(\log \, 4 = 0.60, \, \log \, 5 = 0.69)$
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View SolutionFor the reaction ${N_2}{O_{5(g)}} \to 2N{O_{2(g)}} + \frac{1}{2}{O_{2(g)}}$,the rate constant is $2.3 \times 10^{-2} \ s^{-1}$. Which of the following equations represents the variation of $[{N_2}{O_5}]$ with time?
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