Rate constant of a reaction is `0.0693" min"^(-1)`. Starting with `10" mol L"^(-1)`, rate of reaction after 10 minutes will be

Options

A. `0.0693" M min"^(-1)`
B. `0.0693xx2.5" M min"^(-1)`
C. `0.0693xx5" M min"^(-1)`
D. `0.0693xx10" M min"^(-1)`

Answer

C

Explanation

To calculate concentration after 10 mm :
`k = (2*303)/(t) log""([A]_(0))/([A]) therefore 0*0693 = (2*303)/(10)log ""(10)/([A])`
or `log""(10)/([A]) = 0*3010 or (10)/([A]) =2`
or `[A] = 5 mol L^(-1)`
Rate of reaction after 10 min = k [A]
` = 0*0693 xx 5 mol L^(-1) min^(-1)`