1.

The rate constant for the reaction : 2N_(2) O_(5) rarr 4NO_(2) +O_(2) is 3.0xx10^(-5) "sec"^(-1) . If the rate is 2.40xx10^(-5) mol "litre"^(-1) "sec"^(-1) then the concentration of N_(2)O_(5)(in mol "litre"^(-1) ) is :

Answer»

`1.4`
`1.2`
`0.02`
`0.8`

Solution :(D) RATE`= K[N_(2)O_(5)]`
or `[N_(2)O_(5)] = ("Rate")/(k)`
`= (2.40xx10^(-5))/(3.0xx10^(-5))` =0.8


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