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Tema [17]
3 years ago
7

Integrated rate law for second order unimolecular irreversible

Chemistry
1 answer:
kirill115 [55]3 years ago
5 0

Answer:

The rate law for second order unimolecular irreversible reaction is

\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }

Explanation:

A second order unimolecular irreversible reaction is

2A → B

Thus the rate of the reaction is

v = -\frac{1}{2}.\frac{d[A]}{dt} = k.[A]^{2}

rearranging the ecuation

-\frac{1}{2}.\frac{k}{dt} = \frac{[A]^{2}}{d[A]}

Integrating between times 0 to <em>t </em>and between the concentrations of [A]_{0} to <em>[A].</em>

\int\limits^0_t -\frac{1}{2}.\frac{k}{dt} =\int\limits^A_{0} _A\frac{[A]^{2}}{d[A]}

Solving the integral

\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }

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Thus boron is the element which has properties of both metals and nonmetals.

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Naphthalene, C10H8, melts at 80.2°C. If the vapour pressure of the liquid is 1.3 kPa at 85.8°C and 5.3 kPa at 119.3°C, use th
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(a) One form of the Clausius-Clapeyron equation is

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  • P₂ = 5.3 kPa
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  • T₂ = 119.3°C = 392.46 K

Solving for ΔHv:

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(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:

  • ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
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  • 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
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  • T₂ = 488.1 K = 214.94 °C

(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.

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