Question

.

Answer 1

you have to use the stoichiometric coefficient of \(H_2\).

Assume the overall reaction has a coefficient of 1, so \(H_2\) is reacting 3x as fast as the overall reaction.

Answer 2

Can you show me how I should write that? I'm a little confused.
Would I write the law as Rate=k[N2][H2]^3=[NH3]^2?

Answer 3

I already determined the formation of NH3 which is 4.9 x 10^-2 M/s and the reactant of N2 reacting which is 2.5 x 10^-2. Would I calculate these into the rate law too? I'm confused

Answer 4

for this you dont need the rate law

\(rate=\dfrac{d[reaction]}{dt}=\dfrac{1}{3}\dfrac{d[H_2]}{dt}\)

Answer 5

Oh okay thanks!

Answer 6

np!

Answer 7

aaronq is correct.
Since H2 is a reactant, it is even better to write:

\(rate=\dfrac{d[reaction]}{dt}=-\dfrac{1}{3}\dfrac{d[H_2]}{dt}\)

Answer 8

right! i missed that negative sign, thanks @Vincent-Lyon.Fr

Answer 9

No problem ;-)