Rate of Ammonia Prod. in N2+3H2 Reaction

AI Thread Summary
The discussion centers on the confusion regarding the rate of ammonia production from the reaction of nitrogen and hydrogen, represented by the equation N2 + 3 H2 --> 2 NH3. The stated rate of ammonia production is 5.0 x 10-6 mol/(L * s), leading to questions about the necessity of calculating a "general reaction rate." Participants express uncertainty about the term "general reaction rate," suggesting it may not be clearly defined in the textbook. They agree that the provided rate is sufficient to derive the rates of nitrogen and hydrogen transformation using the same units. Overall, the conversation highlights a lack of clarity in terminology and the interpretation of reaction rates in chemical kinetics.
ThatDude
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Homework Statement



This is not a homework question, rather, I am just trying to clarify a concept and example in my textbook.

Ammonia is produced by the reaction of Nitrogen (N2) with hydrogen (H2) as shown in this equation.

N2 + 3 H2 --> 2 NH3

The rate of ammonia production is 5.0 x 10-6 mol/(L * s)

What is the general reaction rate?
What are the corresponding rates for nitrogen and hydrogen transformation?

2. The attempt at a solution

Calculation:

= (1/2) ( 5.0 x 10-6 mol/(L * s) )

= 2.5 x 10-6 mol/(L * s)

My question is: aren't they already giving us a rate when they say that "The rate of ammonia production is 5.0 x 10-6 mol/(L * s)" ... so why is it necessary to use the general reaction rate law to figure out the "rate". Isn't the rate already given. I'm kind of confused about this.
 
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ThatDude said:

Homework Statement



This is not a homework question, rather, I am just trying to clarify a concept and example in my textbook.

Ammonia is produced by the reaction of Nitrogen (N2) with hydrogen (H2) as shown in this equation.

N2 + 3 H2 --> 2 NH3

The rate of ammonia production is 5.0 x 10-6 mol/(L * s)

What is the general reaction rate?
What are the corresponding rates for nitrogen and hydrogen transformation?

2. The attempt at a solution

Calculation:

= (1/2) ( 5.0 x 10-6 mol/(L * s) )

= 2.5 x 10-6 mol/(L * s)

My question is: aren't they already giving us a rate when they say that "The rate of ammonia production is 5.0 x 10-6 mol/(L * s)" ... so why is it necessary to use the general reaction rate law to figure out the "rate". Isn't the rate already given. I'm kind of confused about this.

You are not the only one. I have never heard of a "general reaction rate". It should be defined in the same book.

They give a rate of ammonia production in some unspecified circumstances in reasonable units, rate of change of molar concentration per sec. You can easily work out the second part of the question in those same units.

I am not aware either of a "general reaction rate law", at least no useful one.
 
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