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Mnenie [13.5K]
3 years ago
7

Why does the reaction proceed more quickly when the surface area is increased?

Chemistry
1 answer:
Gelneren [198K]3 years ago
3 0
When particles collide with the surface of the solid.
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Can someone help with this please
OlgaM077 [116]

Answer:

see explanations

Explanation:

4NH₃(g) + 5O₂(g) => 4NO(g) + 6H₂O(g)

Ci(NH₃) = 3.5mole/4L = 0.875M

Cf(NH₃) = 1.6mole/4L = 0.400M

Rate-1 => Δ[NH₃]/Δt = |(0.400M - 0.875M)/3min| = 0.158M/s

Rate-2 => 6(Δ[NH₃]/Δt) =  4(Δ[H₂O]/Δt) => 6/4(0.158M/s) = 0.237M/s

Rate-3 => 5(Δ[NH₃]/Δt) =  4(Δ[O₂]/Δt) => 5/4(0.158M/s) = 0.237M/s

_________________________________________________________

NOTE: When setting up comparative rate expressions for a given reaction, set the rates expressions as equal then swap coefficient values.  Then solve for rate of interest and substitute givens.      

example: for NH₃ and H₂O

  • set rates expressions equal => Δ[NH₃]/Δt =  Δ[H₂O]/Δt
  • then swap and insert coefficients from given rxn ...
  • solve for rate of interest ...

              4NH₃(g) + 5O₂(g) => 4NO(g) + 6H₂O(g)

              =>  6(Δ[NH₃]/Δt) =  4(Δ[H₂O]/Δt)

              =>  Δ[H₂O]/Δt = 6/4(Δ[NH₃]/Δt) = 6/4(0.237M/s) = 0.237M/s

6 0
3 years ago
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