At a certain moment in the reaction `2N_(2)O_(5)rarr4NO_(2)+O_(2)N_(2)O_(5)` is decomposing at the rate 108 mg `L^(-1)` the production rate of `NO_(2)` is

Options

A. `208 mg L^(-1)s^(-1)`
B. `108 mg L^(-1)s^(-1)`
C. `56 mg L^(-1)s^(-1)`
D. `92 mg L^(-1)s^(-1)`

Answer

B

Explanation

`2N_(2)O_(5) rarr 4NO_(2) + O_(2)`
From reaction stoichiometry, 2 mol of `N_(2)O_(5)` can produce 4 mol of `NO_(2)`, that is, just twice the number. Hence, that rate expression is `-(1)/(1)(d[N_(2)O_(5)])/(d t) = (1)/(4)(d[NO_(2)])/(d t)`
But in the question rate is defined in mass per unit volume per unit time, so converting this rate into moles per unit volume per unit time.
Rate of decomposition of `N_(2)O_(5) = (108 mg L^(-1)s^(-1))/(108 g mol^(-1)) = 1 m mol L^(-1)s^(-1)`
Rate of production of `NO_(2) = (d[NO_(2)])/(d t) = 2 xx {-(d[N_(2)O_(5)])/(d t)} = 2m mol L^(-1)s^(-1)`
Rate of production of `NO_(2) = 2 m mol L^(-1)s^(-1) xx` Molar mass of `N_(2) = 92 mg L^(-1)s^(-1)`