In a reaction, `2Ato "products",` the concentration of A decreases from 0.5 mol `L^(-1) to 0.4 mol L^(-1)` in 10 min.
The rate during this interval is

Options

A. `0.05 "mol"L^(-1)"min"^(-1)`
B. `0.42 """mol"" "L^(-1)"min"^(-1)`
C. `0.005 """mol"" "L^(-1)"min"^(-1)`
D. `0.5 """mol"" "L^(-1)"min"^(-1)`

Answer

C

Explanation

Rate of reaction =`-(1)/(2)(d[A])/(dt)`
`=-(d[A])/(dt)=-([A]_(2)-[A]_(1))/(t_(2)-t_(1))=-(0.4-0.5)/(10-0)=(0.1)/(10)`
`0.01 "mol"L^(-1) "min"^(-1)`
`therefore Rate of reaction =(1)/(2)xx0.01`
`=0.005 "mol" L^(-1) "min"^(-1)`