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Chemistry Β· 2025 Β· 5 marks
CBSE 2025 Β· Region 4 Β· Set 1 Β· Q33
(a) (i) The initial concentration of N_2 O_5 in the first order reaction: N_2 O_5( g ) β 2 NO_2( g )+1/2 O_2( g ) was 1 Β· 2 Γ 10^-2 mol L^-1. Theβ¦
(i)
The initial concentration of $\displaystyle \mathrm{N}_{2} \mathrm{O}_{5}$ in the first order reaction : \[\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \] was $\displaystyle 1 \cdot 2 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$. The concentration of $\displaystyle \mathrm{N}_{2} \mathrm{O}_{5}$ after $\displaystyle 60$ minutes was $\displaystyle 0.2 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$. Calculate the rate constant of the reaction at $\displaystyle 318$ K. $\displaystyle [\log 6=0 \cdot 778]$
(ii)
Account for the following :
(I)
We cannot determine the order of a reaction by taking into consideration the balanced chemical equation.
(II)
A bimolecular reaction may become kinetically of first order under a specified condition.
(i)
The rate of the chemical reaction doubles for an increase of $\displaystyle 10$ K in absolute temperature from $\displaystyle 298$ K. Calculate activation energy ( $\displaystyle \mathrm{E}_{\mathrm{a}}$ ). \[\left[2 \cdot 303 \mathrm{R}=19 \cdot 15 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \log 2=0 \cdot 3\right] \]
(ii)
For a reaction : \[2 \mathrm{H}_{2} \mathrm{O}_{2} \xrightarrow{\mathrm{I}^{-}} 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} \] the proposed mechanism is as given below :
(I)
$\displaystyle \mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{I}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{IO}^{-}$(slow)
(II)
$\displaystyle \mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{IO}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{I}^{-}+\mathrm{O}_{2}$ (fast)
(1)
Write rate law for the reaction.
(2)
Write the overall order and molecularity of the reaction.
Marking-scheme solution
(a)(i)
k =
$\displaystyle 2$β
$\displaystyle 303$
π‘
log
[R] o
[R]
k =
$\displaystyle 2$β
$\displaystyle 303$
$\displaystyle 60$ log
1.$\displaystyle 2$ π₯ $\displaystyle 10$β$\displaystyle 2$
0.$\displaystyle 2$ π₯ $\displaystyle 10$β$\displaystyle 2$
$\displaystyle 2$β
$\displaystyle 303$
$\displaystyle 60$ log $\displaystyle 6$
$\displaystyle 2$β
$\displaystyle 303$
k= $\displaystyle 2.98$ x $\displaystyle 10$ -$\displaystyle 2$ min-$\displaystyle 1$ / $\displaystyle 0.0298$ min-$\displaystyle 1$
(ii)
(I)
Order is determined experimentally.
(II)
If one of the reactants is taken in excess.
(b)(i)
log
π$\displaystyle 2$
π$\displaystyle 1$=
πΈπ
$\displaystyle 2$β
$\displaystyle 303$π
[
π$\displaystyle 1$ β
π$\displaystyle 2$]
log
$\displaystyle 2$π$\displaystyle 1$
π$\displaystyle 1$ =
πΈπ
$\displaystyle 19$β
$\displaystyle 15$ [
$\displaystyle 298$ β
$\displaystyle 308$]
$\displaystyle 0$β
$\displaystyle 3$ =
Ea
$\displaystyle 19$β
$\displaystyle 15$ [
$\displaystyle 298$Γ$\displaystyle 308$]
Ea =
$\displaystyle 0$β
$\displaystyle 3$Γ$\displaystyle 19$β
$\displaystyle 15$Γ$\displaystyle 298$Γ$\displaystyle 308$
Ea= $\displaystyle 52729$ Jmolβ$\displaystyle 1$ or $\displaystyle 52.729$ kJmol-$\displaystyle 1$
(ii)
(1). Rate= k[H2O2] [I-]
(2)
Overall order : $\displaystyle 2$/ Second
Molecularity : $\displaystyle 2$ / Bimolecular
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