CBSE Class 12 Chemistry 2023 Delhi Set 1 Solved Paper

Question : 26 of 35

Marks: +1, -0

SECTION - C

(a) For the reaction

2 {N}_2 {O}_{5({g})} \→4 {NO}_{2({g})}+{O}_{2({g})} \text" at "318 {K}

Calculate the rate of reaction if rate of disappearance of

{N}_2 {O}_{5({g})}

1.4 \×10^{-3} {m} {s}^{-1}

.
(b) For a first order reaction derive the relationship

t_{99 \%}=2 t_{90 \%}

Solution: 👈: Video Solution

(a)

2 {~N}_2 {O}_5 \→4 {NO}_2+{O}_2

{-{1}/{2} {d}[{N}_2 {O}_5]}/{{dt}}=+{{1}/{4} {d}[{NO}_2]}/{{dt}}=+{{d}[{O}_2]}/{{dt}}

=2 \×{1}/{4} \×1.4 \×10^{-3}

[Rate of disappearance

=1.4 \×10^{-3} { ms}^{-1}]

=0.7 \×10^{-3} {ms}^{-1}

(b) For a first order of reaction

t={2.303}/{K} \log {a}/{a-x}

t_{99 \%}={2.303}/{K} \log {100}/{1}

={2.303}/{K} \log 100

={2.303 \×2}/{K}={4.606}/{K}

and

t_{90 \%}={2.303}/{K} \log {a}/{a-x}

={2.303}/{K} \log 10={2.303}/{K}

{t_{99 \%}}/{t_{90 \%}}=2

t_{99 \%}=2 t_{90 \%}

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