Consider the reaction 2 N 2 O 4 leftharpoons 4 NO 2 and given that (-d[ N 2 O 4]/dt)=k and (d[ NO 2]/d t)=k prime, then

Question

Consider the reaction 2 N 2 O 4 leftharpoons 4 NO 2 and given that (-d[ N 2 O 4]/dt)=k and (d[ NO 2]/d t)=k prime, then (A) 2 k prime=k (B)  k prime=2 k (C) k prime=k (D) none of these

Tardigrade Admin · Accepted Answer

2 N 2 O 4 leftharpoons 4 NO 2 Rate =-(1/2) (d[ N 2 O 4]/dt)=+(1/4) (d[ NO 2]/d t) (1/2) k=(1/4) k prime ⇒ k prime=2 k

Q. Consider the reaction $2 N_{2} O_{4} ⇌ 4 N O_{2}$ and given that $\frac{- d [ N _{2} O _{4} ]}{d t} = k$ and $\frac{d [ N O _{2} ]}{d t} = k^{'}$ , then

$2 k^{'} = k$

$k^{'} = 2 k$

$k^{'} = k$

none of these

$2 N_{2} O_{4} ⇌ 4 N O_{2}$

Rate $= - \frac{1}{2} \frac{d [ N _{2} O _{4} ]}{d t} = + \frac{1}{4} \frac{d [ N O _{2} ]}{d t}$

$\frac{1}{2} k = \frac{1}{4} k^{'}$

$\Rightarrow k^{'} = 2 k$

Q. Consider the reaction 2N2​O4​⇌4NO2​ and given that dt−d[N2​O4​]​=k and dtd[NO2​]​=k′, then

2k′=k

k′=2k

k′=k