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Bess [88]
4 years ago
14

The reaction between ethyl bromide (C2H5Br) and hydroxide ion in ethyl alcohol is shown below at 330 K. C2H5Br(alc) + OH −(alc)

→ C2H5OH(l) + Br−(alc) The reaction is first order each in ethyl bromide and hydroxide ion. When [C2H5Br] is 0.0478 M and [OH − ] is 0.162 M, the rate of disappearance of ethyl bromide is 2.8 10-7 M/s.
What is the value of the rate constant?
Chemistry
1 answer:
hram777 [196]4 years ago
3 0

Answer:

3.62 x 10⁻⁵ M⁻¹s⁻¹.

Explanation:

  • Since the reaction is first order each in ethyl bromide and hydroxide ion.
  • <em>The rate of disappearance of ethyl bromide = k[C₂H₅Br][OH⁻],</em>

where, k is the rate constant,

[C₂H₅Br] = 0.0478 M, and [OH⁻] = 0.162 M.

∴ The rate constant (k) = The rate of disappearance of ethyl bromide / [C₂H₅Br][OH⁻] = (2.8 x 10⁻⁷ M/s) / (0.0478 M)(0.162 M) = 3.62 x 10⁻⁵ M⁻¹s⁻¹.

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