1.

The rate constant for the reaction , 2N_(2)O_(5) to 4NO_(2) + O_(2) is 3 xx 10^(-5) sec^(-1) . If the rateis 2.40 xx 10^(-5) mol "litre"^(-1) sec^(-1) . Then the concentration of N_(2)O_(5) (in mol litre^(-1) ) is

Answer»

`1.4`
1.2
`0.04`
`0.8`

Solution :Rate = `k (N_(2)O_(5))` HENCE `2.4 xx 10^(-5) = 3.0 xx 10^(-5) (N_(2)O_(5))`
or `(N_(2)O_(5)) = 0.8` mol `L^(-1)`


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