Question

For the non-stoichiometric reaction

\[\ce{2A + B -> C + D}\], the following kinetic data were obtained in three separate experiments, all at 298 K.

Initial concentration (A)	Initial concentration (B)	Initial rate of formation of C (mol dm−3 s−1)
0.1 M	0.1 M	1.2 × 10−3
0.1 M	0.2 M	1.2 × 10−3
0.2 M	0.1 M	2.4 × 10−3

The rate law for the formation of C is:

पर्याय

• `("d"["C"])/"dt" = "k"["A"]["B"]`

• `("d"["C"])/"dt" = "k"["A"]^2["B"]`

• `("d"["C"])/"dt" = "k"["A"]["B"]^2`

• `("d"["C"])/"dt" = "k"["A"]`

Answer 1

`("d"["C"])/"dt" = "k"["A"]`

Explanation:

The rate law expression is k[A]^x[B]^y

∴ `("d"["C"])/"dt"` = k[A]^x[B]^y

Substituting the given values in above equation for three separate experiments we get,

1.2 × 10⁻³ = k[0.1]^x[0.1]^y ........(i)

1.2 × 10⁻³ = k[0.1]^x[0.2]^y ........(ii)

2.4 × 10⁻³ = k[0.2]^x[0.1]^y ........(iii)

Dividing equation (i) by (ii)

`(1.2 xx 10^-3)/(1.2 xx 10^-3) = ("k"[0.1]^"x"[0.1]"y")/("k"[0.1]^"x"[0.2]^"y")`

1 = `[1/2]^"y"` .....(∴ y = 0)

Now Dividing equation (i) by (iii)

`(1.2 xx 20^-3)/(2.4 xx 10^-3) = ("k"[0.1]^"x"[0.1]^"y")/("k"[0.2]^"x"[0.1]^"y")`

`[1/2]^1 = [1/2]^"x"`

x = 1

Hence, the rate law expression becomes

`("d"["C"])/"dt"` = k[A]¹[B]⁰ = k[A] (∵ [B]⁰ = 1)