Zero order reaction, NEED HELP!

What are the units of K, the rate constant, for zero order reactions?

This may be linked to the question above it

-" In the following balanced reaction, the rate of change of NH3 wasmeasured at -80.0 M/s. What was the rate of change owater at the same time? Reaction = 4 NH3 + 5 O2 --> 4 NO + 6 H2O

-d[A] / dt = k x [A]^n... where n is the order of the reaction..

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if n=0
-d[A] / dt = k x [A]^0 = k x 1 = k

then.. rearranging
d[A] = -k dt

integrating from (time, concentration points of (0, [Ao]) to (t, [At])... ie initial concentration = [Ao] at time = 0 and final concentration = [At] at time = t. and we get

∫d[A] = -k ∫dt

[At] - [Ao] = -k(t - 0)

[At] - [Ao] / t = -k

since [A] is in M.. molarity..
units of K = M/s

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if the reaction was 1st order..
-d[A] / dt = k x [A]¹
1/[A] d[A] = -k dt
ln([At]/[Ao]) = -kt

k has units = ln([At]/[Ao]) / t = 1/sec

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if 2nd order
if the reaction was 1st order..
-d[A] / dt = k x [A]²
1/[A]² d[A] = -k dt
1/[At] - 1/[Ao] = -kt

and units of K = 1/(M x sec)

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for every 4 NH3 consumed, you get 6 H2O's.. and watch the sign... "-" means concentration of that species is decreasing.. + means it's increasing..

so rate water is increasing is + 6/4 x 80 = +120 M/s

rate=K

so M/time

http://www.kentchemistry.com/moviesfiles…

to answer the question just set up a proportion with the coefficients of the reaction

80/ 4 = x/6