Chemistry - Chemical Kinetics Question with Solution | TestHub

It is given that $\Delta [B]=5\times 10^{-3}mol{L}^{-1}$ and $\Delta t=10sec$.

$RateofappearanceofB=\frac{\Delta [B]}{\Delta t}=\frac{5\times 10^{-3}}{10}=5\times 10^{-4}mol{L}^{-1}se{c}^{-1}$

rate of reaction =$-\frac{1}{2}\frac{\Delta [A]}{\Delta t}=\frac{1}{4}\frac{\Delta [B]}{\Delta t}=\frac{\Delta [C]}{\Delta t}$

$$RateofdisappearanceofA=\frac{\Delta [A]}{\Delta t}=\frac{1}{2}\frac{\Delta [B]}{\Delta t}=\frac{1}{2}\times 5\times 10^{-4}=2.5\times 10^{-4}mol{L}^{-1}se{c}^{-1}$$