CBSE 2026 Β· Region 2 Β· Set 1 Β· Q17 Β· 2 marks
The concentration of the reactant is reduced from $\displaystyle 0.6 \mathrm{~mol} \mathrm{~L}^{-1}$ to $\displaystyle 0.2$ $\displaystyle \mathrm{mol} \mathrm{L}^{-1}$ in $\displaystyle 5$ minutes in a first order reaction. Calculate rate constant of the reaction. $\displaystyle (\log 3=0.48)$ $\displaystyle 2$Rate constant k for the first order reaction is $\displaystyle 2.54 \times 10^{-3} \mathrm{~s}^{-1}$. Calculate the time required for three-fourth of the reactant to decompose. \[(\log 4=0.60) \]
The concentration of the reactant is reduced from $\displaystyle 0.6 \mathrm{~mol} \mathrm{~L}^{-1}$ to $\displaystyle 0.2$ $\displaystyle \mathrm{mol} \mathrm{L}^{-1}$ in $\displaystyle 5$ minutes in a first order reaction. Calculate rate constant of the reaction. $\displaystyle (\log 3=0.48)$ $\displaystyle 2$
Rate constant k for the first order reaction is $\displaystyle 2.54 \times 10^{-3} \mathrm{~s}^{-1}$. Calculate the time required for three-fourth of the reactant to decompose. \[(\log 4=0.60) \]
Marking-scheme solution
(A)
k = $\displaystyle 2.303$
π‘
log
[π
]$\displaystyle 0$
[π
]
k= $\displaystyle 2.303$
$\displaystyle 5$ min log $\displaystyle 0.6$
or k= $\displaystyle 2.303$
log $\displaystyle 3$
k = $\displaystyle 2.303$
X $\displaystyle 0.48$
= 0.22minβ$\displaystyle 1$
17.(B)
k = $\displaystyle 2.303$
π‘
log [π
]$\displaystyle 0$
[π
]
t= $\displaystyle 2.303$
π
[π
]$\displaystyle 0$
$\displaystyle 4$ [π
]$\displaystyle 0$
t=
$\displaystyle 2.54$ π $\displaystyle 10$β$\displaystyle 3$ log $\displaystyle 4$
t=
$\displaystyle 2.54$ π $\displaystyle 10$β$\displaystyle 3$ X $\displaystyle 0.60$
= $\displaystyle 5.44$ Γ$\displaystyle 102$ s / $\displaystyle 544$ s
Chemical KineticsIntegrated Rate EquationsApplyvery_short_answermedium
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CBSE Class 12 Chemistry past-paper question from the 2026board exam, with the answer as CBSEβs own marking scheme gives it. Where our answers come from.