For the reaction NO2 + CO → CO2+NO the experimental rate expression is -(dc/dt)=k[NO2]2 the number of molecules of CO involves in the slowest step will be

Question

For the reaction NO2 + CO → CO2+NO the experimental rate expression is -(dc/dt)=k[NO2]2 the number of molecules of CO involves in the slowest step will be (A) 0 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

Since there is no[CO] involved in rate expression therefore CO is involved in fast step.

Q. For the reaction $N O_{2} + CO \to C O_{2} + NO$ the experimental rate expression is $- \frac{d c}{d t} = k [N O_{2}]^{2}$ the number of molecules of $CO$ involves in the slowest step will be

$0$

$1$

$2$

$3$

Since there is no $[CO]$ involved in rate expression therefore $CO$ is involved in fast step.

Q. For the reaction NO2​+CO→CO2​+NO the experimental rate expression is −dtdc​=k[NO2​]2 the number of molecules of CO involves in the slowest step will be

0

1

2

3

Since there is no[CO] involved in rate expression therefore CO is involved in fast step.

Q. For the reaction $N O_{2} + CO \to C O_{2} + NO$ the experimental rate expression is $- \frac{d c}{d t} = k [N O_{2}]^{2}$ the number of molecules of $CO$ involves in the slowest step will be

$0$

$1$

$2$

$3$

Since there is no $[CO]$ involved in rate expression therefore $CO$ is involved in fast step.