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oee [108]
3 years ago
9

In the Haber process for the production of ammonia, what is the relationship between the rate of production of ammonia and the r

ate of consumption of hydrogen?
N2(g) + 3 H2(g) -> 2 NH3(g)

Answer Choices
a.) Δ[NH3] / Δt = −3/2 Δ [H2] / Δt
b.) Δ[NH3] / Δt = Δ [H2] / Δt
c.) 3/2 Δ[NH3] / Δt = Δ [H2] / Δt
d.) Δ[NH3] / Δt = −2/3 Δ [H2] / Δt
e.) Δ[NH3] / Δt = 2/3 Δ [H2] / Δt
f.) 2/3 Δ[NH3] / Δt = −Δ [H2] / Δt
Chemistry
2 answers:
kotykmax [81]3 years ago
5 0

Answer:

Δ[NH₃]/Δt  = 2/3 ( Δ[H₂]/Δt )

Explanation:

From the reaction it can be known that for every 2 moles of ammonia that are produced, 3 moles of hydrogen are consumed. H2 consumption is 3/2 times more than NH3 production. For the ratio, the H2 rate must be decreased by a value of 2/3.

tangare [24]3 years ago
3 0

Answer:

Δ[NH₃]/Δt  = 2/3 ( Δ[H₂]/Δt )

Explanation:

For determining rates as a function of reaction coefficients one should realize that these type problems are <u>always in pairs</u> of reaction components. For the reaction N₂ + 3H₂ => 2NH₃ one can compare ...

Δ[N₂]/Δt  ∝ Δ[H₂]/Δt, or

Δ[N₂]/Δt  ∝ Δ[NH₃]/Δt, or

Δ[H₂]/Δt  ∝ Δ[NH₃]/Δt, but never 3 at a time.

So, set up the relationship of interest ( ammonia rate vs. hydrogen rate)... nitrogen rate is ignored.

Δ[H₂]/Δt  ∝ Δ[NH₃]/Δt

Now, 'swap' coefficients of balanced equation and apply to terms given then set term multiples equal ...

N₂ + 3H₂ => 2NH₃   => 2(Δ[H₂]/Δt)  = 3(Δ[NH₃]/Δt) => 2/3(Δ[H₂]/Δt)  = (Δ[NH₃]/Δt)

NOTE => Comparing rates individually of the component rates in reaction process, the rate of H₂(g) consumption is 3/2 times <u>faster</u> than NH₃(g) production (larger coefficient).  So, in order to compose an equivalent mathematical relationship between the two, one must reduce the rate of the H₂(g) by 2/3 in order to equal the rate of NH₃(g) production. Now, given the rate of one of the components as a given, substitute and solve for the unknown.

CAUTION => When Interpreting rate of reaction one should note that the  rate expression for an individual reaction component defines 'instantaneous' rate or speed. <u>This means velocity (or, speed) does not have 'signage'</u>. Yes, one may say the rate is higher or lower as time changes but that change is an acceleration or deceleration for one instantaneous velocity to another. Acceleration and Deceleration do have signage but the positional instantaneous velocity (defined by a point in time) does not. Thus is reason for the  'e-choice' answer selection without the signage associated with the expression terms.  

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