In the reversible reaction, 2NO2 xrightleftharpoons[k2]k1N2O4 , the rate of disappearance of NO2 is equal to

Question

In the reversible reaction, 2NO2 xrightleftharpoons[k2]k1N2O4 , the rate of disappearance of NO2 is equal to (A)  (2k1/k2)[NO2]2  (B)  2k1 [NO2]-2k2[N2O4]  (C)  2k1 [NO2]2-2k2[N2O4]  (D)  (2k1 - k2) [NO2]

Tardigrade Admin · Accepted Answer

2NO2 xrightleftharpoons[k2]k1 N2O4 The rate of disappearance of NO2 = (-1/2) (d[NO2]/dt) ⇒ (-1/2) (d[NO2]/dt) = k1[NO2]2 - k2[N2O4] (-dNO2/dt) = 2k1[NO2]2 - 2k2[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

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

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

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

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

$2 N O_{2} k_{1} k_{2} N_{2} O_{4}$
The rate of disappearance of $N O_{2} = \frac{- 1}{2} \frac{d [ N O _{2} ]}{d t}$
$\Rightarrow \frac{- 1}{2} \frac{d [ N O _{2} ]}{d t} = k_{1} [N O_{2}]^{2} - k_{2} [N_{2} O_{4}]$
$\frac{- d N O _{2}}{d t} = 2 k_{1} [N O_{2}]^{2} - 2 k_{2} [N_{2} O_{4}]$

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​]−2k2​[N2​O4​]

2k1​[NO2​]2−2k2​[N2​O4​]

(2k1​−k2​)[NO2​]