Q.
$A+2B \rightarrow C$ , the rate equation for the reaction is given as
Rate $=k\left[\right. A \left]\right.\left[\right. B \left]\right.$
If the concentration of A is kept the same but that of B is doubled what will happen to the rate itself?
NTA AbhyasNTA Abhyas 2022
Solution:
The order of the overall reaction is 2.
Rate $=k\left[\right.A\left]\right.\left[\right.B\left]\right.=R$
$R^{'}=k\left[\right.A\left]\right.\left[\right.2B\left]\right.$
$\frac{R}{R^{'}}=\frac{k \left[\right. A \left]\right. \left[\right. B \left]\right.}{k \left[\right. A \left]\right. \left[\right. 2 B \left]\right.}=\frac{k \left[\right. A \left]\right. \left[\right. B \left]\right.}{2 k \left[\right. A \left]\right. \left[\right. B \left]\right.}$
$\Rightarrow 2R=R^{'}$ i.e, rate become doubles.
Rate $=k\left[\right.A\left]\right.\left[\right.B\left]\right.=R$
$R^{'}=k\left[\right.A\left]\right.\left[\right.2B\left]\right.$
$\frac{R}{R^{'}}=\frac{k \left[\right. A \left]\right. \left[\right. B \left]\right.}{k \left[\right. A \left]\right. \left[\right. 2 B \left]\right.}=\frac{k \left[\right. A \left]\right. \left[\right. B \left]\right.}{2 k \left[\right. A \left]\right. \left[\right. B \left]\right.}$
$\Rightarrow 2R=R^{'}$ i.e, rate become doubles.