1.

For the non-stoichiometric reaction: 2A + B to C +D, the following kinetic data were obtained in three separate experiments, all 298 K The rate law for the formation of C is:

Answer»

`(dC)/(dt) = K[A]`
`(dC)/(dt) = k[A][B]`
`(dC)/(dt) = k[A]^(2)[B]`
`(dC)/(dt) = k[A][B]^(2)`

Solution :a) For the reaction, `2A + B to C + D`
Rate of reaction,
`-1/2(d[A])/(dt) = (-d[B])/(dt) = (d[C])/(dt) = (d[D])/(dt)`
Now, rate of reaction, `(d[C])/(dt) = k[A]^(X)[B]^(y)`
From table,
`1.2 xx 10^(-3)= k(0.1)^(x)(0.1)^(y)`…………(i)
`1.2 xx 10^(-3)=k(0.1)^(x)(0.2)^(y)`..........(ii)
`2.4 xx 10^(-3)=k(0.2)^(x)(0.1)^(y)`............(III)
On dividing equation i) by ii), we GET
`(1.2xx10^(-3))/(1.2xx10^(-3)) = (k(0.1)^(x)(0.1)^(y))/(k(0.1)^(x)(0.2)^(y))`
`1=(1/2)^(y)` or `(1)^(@) = (1/2)^(Y)`therefore y=0
On dividing equation (i) by (ii), we get
`(1.2 xx 10^(-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)` or x=1
Hence, `(d[C])/(dt) = k[A]^(1)[B]^(0)=k[A]`


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