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For the reaction at 273 K NO(g) + O(3)...

For the reaction at `273 K`
`NO(g) + O_(3)(g) rarr NO_(2)(g) + O_(2)(g)`
It is observed that the pressure of `NO(g)` falls form `700 mm Hg` to `500 mm Hg` in `250 s`. Calculate the average rate of reaction in (a) `atm s^(-1)` , (b) `mol L^(-1) s^(-1)`.

Text Solution

Verified by Experts

The correct Answer is:
1.053
`NO(g) + O_(3)(g) to NO_(2)(g) + O_(2)(g)`
Average rate `=("Change of pressure")/("Change of time") = (DeltaP)/I(Deltat)`
`=(700-500)/250 = 200/250`
`=200/(760 xx 250) atm s^(-1) = 1.053 xx 10^(-3) atm s^(-1)`
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