In the reversible reaction, 2 NO 2 -[k1][K2] N 2 O 4 the rate of disappearance of N O2 is equal to

Question

In the reversible reaction, 2 NO 2 ->[k1][K2] N 2 O 4 the rate of disappearance of N O2 is equal to (A) (2k1/k2)[NO2]2 (B) 2k1[NO2]2 - 2k2[N2O4] (C) 2k1[NO2]2 - k-1[N2O4] (D) (2k1 - k2) [NO2]

Tardigrade Admin · Accepted Answer

2NO2 ->[k1][k2] N2O4 Rate = -(1/2) (d[NO2]/dt) = k1[NO2]2 - k2[N2O4] ∴ Rate of disappearance of NO2 i.e., -(d[NO2]/dt) = 2k1[NO2]2 - k2[N2O4]

Q. In the reversible reaction,
$2 N O_{2} k_{1} K_{2} N_{2} O_{4}$
the rate of disappearance of $N O_{2}$ is equal to

$\frac{2 k _{1}}{k _{2}} [N O_{2}]^{2}$

$2 k_{1} [N O_{2}]^{2} - 2 k_{2} [N_{2} O_{4}]$

$2 k_{1} [N O_{2}]^{2} - k_{- 1} [N_{2} O_{4}]$

$(2 k_{1} - k_{2}) [N O_{2}]$

$2 N O_{2} k_{1} k_{2} N_{2} O_{4}$
Rate = $- \frac{1}{2} \frac{d [ N O _{2} ]}{d t} = k_{1} [N O_{2}]^{2} - k_{2} [N_{2} O_{4}]$
$∴$ Rate of disappearance of $N O_{2}$ i.e.,
$- \frac{d [ N O _{2} ]}{d t} = 2 k_{1} [N O_{2}]^{2} - k_{2} [N_{2} O_{4}]$

Q. In the reversible reaction, 2NO2​k1​K2​​N2​O4​ the rate of disappearance of NO2​ is equal to

k2​2k1​​[NO2​]2

2k1​[NO2​]2−2k2​[N2​O4​]

2k1​[NO2​]2−k−1​[N2​O4​]

(2k1​−k2​)[NO2​]