Question

The reaction $2 NO+Br_{2}$ $\rightarrow$ $2NOBr$, obeys the following mechanism :
$NO+Br_{2}$ $\ce{ <=>[Fast] }$ $NOBr_{2} ;+NO$ $\ce{ ->[Slow] }$ $2NOBr$ The rate expression of the above reaction can be written as

• A. $r=k\left[NO\right]^{2}$ $\left[Br_{2}\right]$
• B. $r=k\left[NO\right]$ $\left[Br_{2}\right]$
• C. $r=k {NO^2Br_2}$
• D. $r=k\left[NOBr_{2}\right]$

Answer 1

Answer: (A)

For slowest step : rate $=k\left[NOBr_{2}\right]\left[NO\right]$ $\ldots\left(i\right)$
For equilibrium also $ K_{c}=\frac{\left[NOBr_{2}\right]}{\left[NO\right]\left[Br_{2}\right]}$ $\quad$ $\ldots\left(ii\right)$
By eqs. $\left(i\right) and \left(ii\right)$, $r=k\cdot$ $k_{c}\left[NO\right]^{2}\left[Br_{2}\right]$
Rate$=k\left[NO\right]^{2}$ $\left[Br_{2}\right]$