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

Answer:

$1.2\times 10^{2}$ mmol of $NH_{3}$ is left after 20 s.

Explanation:

Initial concentration of $NH_{3}$ = $\frac{150\times 10^{-3}}{450}\times 10^{3}$ 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, $[NH_{3}]=(-0.0038M.s^{-1}\times 20s)+0.333M$

   or, $[NH_{3}]$ = 0.257 M

So, number of mol of $NH_{3}$ left after 20 s = $\frac{0.257}{1000}\times 450$ mol = 0.11565 mol

So, number of mmol of $NH_{3}$ left after 20 s = 115.65 mmol = $1.2\times 10^{2}$ mmol (2 sig. digits)