$$A + 2B \to C,$$   the rate equation for this reaction is given as Rate $$ = K\left[ A \right]\left[ B \right].$$   If the concentration of $$A$$ is kept the same but that of $$B$$  is doubled what will happen to the rate itself ?

Solution :
$$\eqalign{ & {\text{Rate}} = k\left[ A \right]\left[ B \right] = R \cr & R' = k\left[ A \right]\left[ {2B} \right] \cr & \frac{R}{{R'}} = \frac{{k\left[ A \right]\left[ B \right]}}{{k\left[ A \right]\left[ {2B} \right]}} \cr & = \frac{{k\left[ A \right]\left[ B \right]}}{{2k\left[ A \right]\left[ B \right]}} \cr & \Rightarrow 2R = R'\,\,i.e.,\,{\text{rate become doubles}}{\text{.}} \cr} $$