The rate constant k, for the reaction N 2 O 5(g) longrightarrow 2 NO 2(g)+(1/2) O 2(g) is 2.3 × 10-2 s -1. Which equation given below describes the change of [ N 2 O 5] with time? [ N 2 O 5]0 and [ N 2 O 5]t correspond to concentration of N 2 O 5 initially and at time t.

Question

The rate constant k, for the reaction N 2 O 5(g) longrightarrow 2 NO 2(g)+(1/2) O 2(g) is 2.3 × 10-2 s -1. Which equation given below describes the change of [ N 2 O 5] with time? [ N 2 O 5]0 and [ N 2 O 5]t correspond to concentration of N 2 O 5 initially and at time t. (A) [ N 2 O 5]t=[ N 2 O 5]0+k t (B) [ N 2 O 5]0=[ N 2 O 5]t ek t (C)  log [ N 2 O 5]t= log [ N 2 O 5]0+k t (D)  ln ([ N 2 O 5]0/[ N 2 O 5]8)=k t

Tardigrade Admin · Accepted Answer

From the unit of rate constant ( sec -1) it is clear that the reaction is of first order. For Ist order reaction k=(1/t) log e (a/a-x) or k t= log e ([ N 2 O 5]0/[ N 2 O 5] t) k propto (1/t) so, k propto (1/ sec )= s -1

$[N_{2} O_{5}]_{t} = [N_{2} O_{5}]_{0} + k t$

$[N_{2} O_{5}]_{0} = [N_{2} O_{5}]_{t} e^{k t}$

$lo g [N_{2} O_{5}]_{t} = lo g [N_{2} O_{5}]_{0} + k t$

$ln \frac{[ N _{2} O _{5} ] _{0}}{[ N _{2} O _{5} ] _{8}} = k t$

From the unit of rate constant $(se c^{- 1})$ it is clear that the reaction is of first order. For Ist order reaction $k = \frac{1}{t} lo g_{e} \frac{a}{a - x}$
or $k t = lo g_{e} \frac{[ N _{2} O _{5} ] _{0}}{[ N _{2} O _{5} ] t}$
$k \propto \frac{1}{t}$ so,
$k \propto \frac{1}{s e c} = s^{- 1}$

Q. The rate constant k, for the reaction N2​O5​(g)⟶2NO2​(g)+21​O2​(g) is 2.3×10−2s−1. Which equation given below describes the change of [N2​O5​] with time? [N2​O5​]0​ and [N2​O5​]t​ correspond to concentration of N2​O5​ initially and at time t.

[N2​O5​]t​=[N2​O5​]0​+kt

[N2​O5​]0​=[N2​O5​]t​ekt

log[N2​O5​]t​=log[N2​O5​]0​+kt

ln[N2​O5​]8​[N2​O5​]0​​=kt

From the unit of rate constant (sec−1) it is clear that the reaction is of first order. For Ist order reaction k=t1​loge​a−xa​ or kt=loge​[N2​O5​]t[N2​O5​]0​​ k∝t1​ so, k∝sec1​=s−1

$[N_{2} O_{5}]_{t} = [N_{2} O_{5}]_{0} + k t$

$[N_{2} O_{5}]_{0} = [N_{2} O_{5}]_{t} e^{k t}$

$lo g [N_{2} O_{5}]_{t} = lo g [N_{2} O_{5}]_{0} + k t$

$ln \frac{[ N _{2} O _{5} ] _{0}}{[ N _{2} O _{5} ] _{8}} = k t$

From the unit of rate constant $(se c^{- 1})$ it is clear that the reaction is of first order. For Ist order reaction $k = \frac{1}{t} lo g_{e} \frac{a}{a - x}$
or $k t = lo g_{e} \frac{[ N _{2} O _{5} ] _{0}}{[ N _{2} O _{5} ] t}$
$k \propto \frac{1}{t}$ so,
$k \propto \frac{1}{s e c} = s^{- 1}$