1.

For the reaction N_(2)O_(5) (g) to 2NO_(2) (g) + (1)/(2) O_(2) (g) the value of rate of disappearance of N_(2)O_(5) is given as 6.25 xx 10^(-3) mol L^(-1) S^(-1) . The rate of formation of NO_(2) and O_(2) is given respectively as

Answer»

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

Solution :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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