Reaction rate : At some time, we observe that the reaction
`2N_(2)O_(5)(g)rarr 4NO_(2)(g)+O_(2)(g)`
is forming `NO_(2)` at the rate of `0.0072 mol L^(-1)s^(-1)`.
(i) What is the rate of reaction at this time ?
(ii) What is the rate of formation of `O_(2)` at this time ?
(iii) What is the rate of consumption of `N_(2)O_(5)` at this time ?
Strategy : For expressing the rate of such reaction where stoichiometric coefficents of reactants or products are not equal to one, rate of any of the products is divided by their respective stoichiometric coefficients. Use the mole ratios from the balanced equation to determine the rates of change of other products and reactants.

Explanation

(i) The rate of rection can be calculated from the rate of decreas of any reactant concentration or the rate increase of any product concentration :
Averag rate of reaction
`= -(1)/(2) ((Delta[N_(2)O_(5)])/(Delta t))= 1/4((Delta[NO_(2)])/(Deltat))`
`= (1)/(1) ((Delta[O_(2)])/(Delta t))`
Average rate of formation of `NO_(2)`
`= ((Delta[NO_(2)])/(Delta t)) = 0.0072 mol L^(-1)s^(-1)`
Therefore, average reaction rate
`= (1)/(4) ((Delta[NO_(2)])/(Delta t))`
`=(1)/(4) (0.0072mol L^(-1)s^(-1))`
`= 0.0018mol L^(-1)s^(-1)`
Remember that the mol in thisd unit is interpreted as "mol of reaction".
(ii) The balanced equation equation gives the reaction ratio
`(1mol O_(2))/(4mol NO_(2))`
Thus, rate of formation of `O_(2)`
`= (Delta[O_(2)])/(Delta t)=((Delta[NO_(2)])/(Delta t)) ((1mol O_(2))/(4mol NO_(2))`
`= ((0.0072mol NO_(2))/(L.s)) ((1mol O_(2))/(4mol NO_(2)))`
`(0.0018mol O_(2))/(L.s)`
Alternatively, according to general rate of reaction
`(1)/(1) ((Delta[O_(2)])/(Delta t))= (1)/(4) ((Delta[NO_(2)])/(Delta t))`
Thus,
`(Delta[O_(2)])/(Delta t) = (1)/(4)((0.0072mol)/(L.s))`
`= 0.0018 mol L^(-1) s^(-1)`
(iii) The balance equation show that 2 mol `N_(2)O_(5)` is consumed for every 4 mol `NO_(2)` that is formed. Because `[N_(2)O_(5)]` is decreasing as `[NO_(2)]` increases, we should write reaction ratio as
`-(2mol N_(2)O_(5))/(4mol NO_(2))`
Thus, rate of consumption of `N_(2)O_(5)`
`= - (Delta[N_(2)O_(5)])/(Delta t)= - ((0.0072mol NO_(2))/(L.s))xx -(2mol N_(2)O_(5))/(4mol NO_(2))`
`= 0.0036 (mol N_(2)O_(5))/(L.s)`
Alternatively, according to general rate of reaction
`-(1)/(2) ((Delta[N_(2)O_(5)])/(Delta t))= (1)/(4) ((Delta[NO_(2)])/(Delta t))`
Thus,
`- (Delta[N_(2)O_(5)])/(Delta t) =(2)/(4) (0.0072 mol L^(-1) s^(-1))`
`= 0.0036 mol L^(-1)s^(-1)`