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The reaction 2N2O5to4NO2+O2 forms NO2 at...

The reaction `2N_2O_5to4NO_2+O_2` forms `NO_2` at the rate of 0.0072 mol `L^(-1)s^(-1)` after a certain time.
What is the rate of change of `[N_2O_5]` at this time ?

Text Solution

Verified by Experts

The rate of the reaction is expressed as
Rate = `-1/2 d[N_2O_5]//dt=1/4 d[NO_2]//dt=d[O_2]//dt` and given
`2 d [NO] //dt = 0.0072 mol L^(-1) s^(-1)`
Rate of disappearance of `N_2O_5= 1/2 xx" rate of appearance of "NO_2`
`-d[N_2O_5] // dt= 1/2 d [NO]//dt = 1/2 xx 0.0072 = -0.036" mole"//L//s`.
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