Question

The rate law for the reaction 2A + B →C is –rA= kACA2CBwithkA= 25 (L/mol)2sec. What arekBandkC?

Answer 1

Step-by-step explanation:

The given reaction equation is as follows.

`2A + B \rightarrow C`

So, rate constants for different reactants and products written as follows.

`\frac{-k_(A)}{\text{stoichiometric coefficient of A}} = \frac{-k_(B)}{\text{stoichiometric coefficient of B}} = \frac{k_(C)}{\text{stoichiometric coefficient of C}}`

As per the reaction equation, the stoichiometric coefficients of reactants and products are as follows.

A = -2

B = -1

C = 1

Therefore,

`\frac{-k_(A)}{\text{stoichiometric coefficient of A}} = \frac{-k_(B)}{\text{stoichiometric coefficient of B}} = \frac{k_(C)}{\text{stoichiometric coefficient of C}}`

`(-k_(A))/(-2) = (-k_(B))/(-1) = (k_(C))/(1)`

`(-k_(A))/(-2) = k_(B) = k_(C)`

Hence,
`k_(B) = k_(C)` =
`(25)/(2) (L/mol)^(2)`

= 12.5
`(L/mol)^(2)`

Thus, we can conclude that
`k_(B)` and
`k_(C)` are 12.5
`(L/mol)^(2)`.