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For the reaction N(2)O(5) rarr 2NO(2) + ...

For the reaction `N_(2)O_(5) rarr 2NO_(2) + (1)/(2) O_(2)`, the rate of disappearance of `N_(2)O_(5)` is `6.25 xx 10^(-3) "mol L"^(-1) s^(-1)`. The rate of formation of `NO_(2)` and `O_(2)` will be respectively.

A

`1.25 xx 10^(-2)` mol `L^(-1) S^(-1)` and ` 6.25 xx 10^(-3)` mol `L^(-1) S^(-1)`

B

`6.25 xx 10^(-3)` mol `L^(-1) S^(-1)` and ` 6.25 xx 10^(-3)` mol `L^(-1) S^(-1)`

C

`1.25 xx 10^(-2)` mol `L^(-1) S^(-1)` and ` 3.125 xx 10^(-3)` mol `L^(-1) S^(-1)`

D

`6.25 xx 10^(-2)` mol `L^(-1) S^(-1)` and ` 3.125 xx 10^(-3)` mol `L^(-1) S^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
Given `(-d[N_(2)O_(5)])/(dt) = 6.25 xx 10^(-3)` mol `L^(-1) S^(-1)`
Fore the reaction , `N_(2)O_(5) to 2NO_(2) + (1)/(2) O_(2)`
`(-d[N_(2)O_(5)])/(dt) = (1)/(2) (d[NO_(2)])/(dt) = (2d[O_(2)])/(dt)`
`therefore (d[NO_(2)])/(dt) = -(2d[N_(2)O_(5)])/(dt) = 1.25 xx 10^(-2)` mol `L^(-1) S^(-1)`
`therefore (d[O_(2)])/(dt) = - (1)/(2) (d[N_(2)O_(5)])/(dt) = 3.125 xx 10^(-3) ` mol `L^(-1) S^(-1)`.
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