Question

The elementary reaction A+B→2C + D The reactor is initially charged with CBo/CAD-3.0, and CAD-0.01 moVL, calculate the value of reaction rate constant, k after 10 min of reaction when the conversion of A is 50 %

Answer 1

Step-by-step explanation:

As the given reaction is as follows.

`A + B \rightarrow 2C + D`

at t = 0
`C_(Ao)`
`3C_(Ao)` 0 0

at t = t'
`C_(Ao) - C_(Ao)X_(Ao)`
`3C_(Ao) - C_(Ao)X_(A)`
`2X_(A)C_(Ao)`
`C_(Ao)X_(A)`

`-r_(A)` =
`(dC_(A))/(dt)` =
`kC_(A)C_(B)`

=
`kC_(Ao)(1 - X_(A))C_(Ao)(3 - X_(A))`

=
`kC^(2)_(Ao)(1 - X_(A))(3 - X_(A))`

`(dC_(Ao)(1 - X_(A)))/(dt)` =
`kC^(2)_(Ao)(1 - X_(A))(3 - X_(A))`

`-C_(Ao)(dX_(A))/(dt)` = kC^{2}_{Ao} (1 - X_{A})(3 - X_{A})[/tex]

`\int_(0)^(0.5)(dX_(A))/((1 - X_(A))(3 - X_(A)))` =
`kC_(Ao)\int_(0)^(10 min)dt`

`\int_(0)^(0.5)[(1)/(2(1 - X_(A))) - (1)/(2(3 - X_(A)))] dX_(A)` =
`kC_(Ao) * (10 - 0)`

`(1)/(2)[-ln (1 - X_(A))]^(0.5)_(0) + (1)/(2)[ln(3 - X_(A))]^(0.5)_(0)` =
`kC_(Ao) * 10`

`(1)/(2)[ln 1 - ln 0.5]^(0.5)_(0) + (1)/(2) * ((2.5)/(3))` =
`kC_(Ao) * 10`

0.2554 =
`kC_(Ao) * 10`

Given,
`C_(Ao)` = 0.01 mol/L

k =
`(0.2554 L mol^(-1) min^(-1))/(0.01 * 10)`

= 2.554
`L mol^(-1) min^(-1)`

Thus, we can conclude that value of given reaction rate constant is 2.554
`L mol^(-1) min^(-1)`.