1.

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

Answer»

6
8
3
2

Solution :`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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