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Zigmanuir [339]
3 years ago
15

The rate constant for this first‑order reaction is

Chemistry
1 answer:
jenyasd209 [6]3 years ago
5 0

Answer:

t = 5.7634 s

Explanation:

  • A → Pdts
  • - rA = K (CA)∧α = - δCA/δt

∴ T = 400°C

∴ α = 1 ....first-order

∴ CAo = 0.950 M

∴ CA = 0.300 M

⇒ t = ?

⇒ - δCA/δt = K*CA

⇒ - ∫δCA/CA = K*∫δt

⇒ Ln (CAo/CA) = K*t

⇒ t = Ln(CAo/CA) / K

⇒ t = (Ln(0.950/0.300)) / (0.200 s-1)

⇒ t = 1.1527 / 0.200 s-1

⇒ t = 5.7634 s

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Answer :

(1) The frequency of photon is, 3\times 10^{10}Hz

(2) The energy of a single photon of this radiation is 1.988\times 10^{-23}J/photon

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Explanation : Given,

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(1) Now we have to calculate the frequency of photon.

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Now put all the given values in the above formula, we get:

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\nu=3\times 10^{10}s^{-1}=3\times 10^{10}Hz    (1Hz=1s^{-1})

The frequency of photon is, 3\times 10^{10}Hz

(2) Now we have to calculate the energy of photon.

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Now put all the given values in the above formula, we get:

E=(6.626\times 10^{-34}Js)\times (3\times 10^{10}s^{-1})

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The energy of a single photon of this radiation is 1.988\times 10^{-23}J/photon

(3) Now we have to calculate the energy in J/mol.

E=1.988\times 10^{-23}J/photon

E=(1.988\times 10^{-23}J/photon)\times (6.022\times 10^{23}photon/mol)

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The energy of an Avogadro's number of photons of this radiation is, 11.97 J/mol

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