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The data given below are for the reactio...

The data given below are for the reaction of NO and `Cl_(2)` to form NOCl at 295 K
`[Cl_(2)]" "[NO]" ""initial rate(mol litre"^(-1)"sec"^(-1))`
`0.05" "0.05" "1xx10^(-3)`
`0.15" "0.05" 3x10^(-3)`
`0.05" "0.15" "9xx10^(-3)`
The rate constant of the reaction is

A

6

B

8

C

3

D

2

Text Solution

Verified by Experts

The correct Answer is:
B
`NO+1/2Cl_(2)toNOCl,r=K[NO]^(x)[Cl_(2)]^(y)`
`1xx10^(-3)=K(0.05)^(x).(0.05)^(y)`……1, `3xx10^(-3)=K(0.05)^(x).(0.15)^(y)`………2
`9xx10^(3)=K.(0.15)^(x).(0.05)^(y)`……….3, `((3))/((1))implies((9xx10^(-3))/(1xx10^(-3)))=(Kxx(0.15)^(x)xx(0.05)^(y))/(Kxx(0.05)(0.05))`
`3^(2)=3^(x)impliesx=2,((2))/((1))implies((3xx10^(-3))/(1xx10^(-3)))=(Kxx(0.05)^(x)xx(0.15)^(y))/(Kxx(0.05)(0.05))`
`3^(1)=3^(y)impliesy=1,r=K.[NO]^(2)[Cl_(2)]^(1),K=8`
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