Question

Under certain conditions the rate of this reaction is zero order in ammonia with a rate constant of ·0.0038Ms−1: 2NH3(g)→N2(g)+3H2(g) Suppose a 450.mL flask is charged under these conditions with 150.mmol of ammonia. How much is left 20.s later? You may assume no other reaction is important. Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.

Answer 1

`1.2* 10^(2)` mmol of
`NH_(3)` is left after 20 s.

Step-by-step explanation:

Initial concentration of
`NH_(3)` =
`(150* 10^(-3))/(450)* 10^(3)` M = 0.333 M

The integrated rate law for the given zero order reaction:

`[NH_(3)]=-kt+[NH_(3)]_(0)`

where,
`[NH_(3)]` represents concentration of
`NH_(3)` after "t" time, k is rate constant and
`[NH_(3)]_(0)` is initial concentration of
`NH_(3)`.

Here, k = 0.0038 M/s,
`[NH_(3)]_(0)` = 0.333 M and t = 20 s

So,
`[NH_(3)]=(-0.0038M.s^(-1)* 20s)+0.333M`

or,
`[NH_(3)]` = 0.257 M

So, number of mol of
`NH_(3)` left after 20 s =
`(0.257)/(1000)* 450` mol = 0.11565 mol

So, number of mmol of
`NH_(3)` left after 20 s = 115.65 mmol =
`1.2* 10^(2)` mmol (2 sig. digits)