Homework Help 15.1

[Andrew Uesugi 3I]

I'm a bit confused with this question.

So the question states:

Complete the following statements relating to the production of ammonia by the Haber process, for which the overall reaction is N2 + 3H2 -> 2 NH3

a) the rate of consuption of N2 is _____ times the rate of consumption of H2.

When I first saw this problem, I thought I had to use the unique average rate. But to my surprise, the solution manual is telling me it's:

rate (n2) = rate(h2) x (1 mol N2 / 3 mol h2) = 1/3 x rate(H2)

I've never seen this in Chapter 15. Can someone explain to me what exactly call for this situation to use what the solution manual stated?

[Marie_Bae_3M]

I'm not entirely sure but I used the method of initial rates: -1/a*(d[A]/dt)=1/b*(d[B]/dt)=1/c*(d[C]/dt) for aA ------> bB+cC. In this case, I did:
-1/1*(d[N2]/dt)=-1/3(d[H2]/dt)
Cancel out the dt's and the -1's.
Then you are left with d[N2] (which is the rate of the concentration on N2) = 1/3*d[H2] (the rate of concentration of H2)

[Wendy_Liu_3A]

If you use the unique average rate law, the result should still be the same.
-d[A]/dt = -1/3*d[B]/dt
d[A]/dt = 1/3*d[B]/dt
In this case, A is N2 and B is H2.
So the rate of consumption of N2 is 1/3 * rate of consumption of H2
The way the textbook does is more conceptual, which was what I did. You can think of it solely in terms of their coefficient.
N2 + 3H2 -> 2 NH3
For every 3 moles of H2 are consumed, 1 mol N2 is consumed. The rate of consumption of H2 is 3 times that of N2. In other words, the rate of consumption of N2 is 1/3 times of H2.

[Helen Shi 1J]

Can someone explain part b and c too?

[Sandhya Rajkumar 1C]

For part b, the balanced equation shows that for every 2 moles of NH3 formed, 3 moles of H2 are consumed. This means that the rate of formation of NH3 would be 2/3 of the rate of consumption of H2. We do the same thing for part c. For every 2 moles of NH3 formed, 1 mole of N2 is consumed. This means that the rate of formation of NH3 is 2 times the rate of consumption of N2. The solution manual uses the ratios of the stoichiometric coefficients to calculate the answers.