Consider the reaction,
` aA +bB to cD.`
if 4M of A are allowed to react with 2 Mof B and concentration of A after 4 seconds is 3 M, calculate the rate of reaction. Mention it in terms of A as well as D.

Explanation

We know that Rate of reaction
`=(" Number of moles per litre of reactant consumed")/("Time taken ") `
Given : Time=4-0= 4 seconds
Number of moles of A consumed = 4M - 3M = 1M
Substituting the values, we get rate of reaction
` (-(d[A] )/( dt)) = (1M )/( 4s ) = 0.25 Ms^(-)`
To find rate of reaction in terms of D. We know that
`-(1)/(a) (d[A])/(dt) = + (1)/(d) (d[D])/( dt)`
` (d[D])/(dt)=- (d)/(a) (d[A])/( dt) = (d)/(a) xx 0.25 Ms^(-)`