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IRISSAK [1]
4 years ago
10

Assuming the relative rate of secondary hydrogen atom abstraction for the chlorination of propane is 3.9 times faster than the r

ate of primary hydrogen abstraction, calculate the relative amounts of propyl chloride and isopropyl chloride obtained in the free radical chlorination of propane. Based on the calculation what percentage of the mixture is expected to be 1−propylchloride?
Chemistry
1 answer:
Ghella [55]4 years ago
7 0

Answer:

% of n-propyl chloride = 43.48 %

Explanation:

There are 2 secondary hydrogens and 6 primary hydrogens

The rate of abstraction of seondary hydrogen = 3.9 X rate of abstraction of primary hydrogen

probability of formation of isopropyl chloride = 3.9 X 1 (relative rate X relative number of secondary hydrogens)

Probability of formation of n-propyl chloride = 1 X 3 (relative rate X relative number of primary hydrogens)

Total probability = 3.9

% of n-propyl chloride = 3 X 100 / 6.9 = 43.48 %

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Suppose that the volume of a particular sample of Cl2 is 718 mL at 675 mmHg and 48C at what temperature in C will the volume be
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Answer:

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Step-by-step explanation:

This looks like a case where we can use the <em>Combined Gas Law</em> to calculate the temperature.

p₁V₁/T₁ = p₂V₂/T₂               Multiply both sides by T₂

p₁V₁T₂/T₁ = p₂V₂                 Multiply each side by T₁

p₁V₁T₂ = p₂V₂T₁                   Divide each side by p₁V₁

T₂ = T₁ × p₂/p₁ × V₂/V₁  

=====

Data:

We must convert the pressures to a common unit. I have chosen atmospheres.

p₁ = 675 mmHg × 1atm/760 mmHg = 0.8882 atm

V₁ = 718 mL = 0.718 L

T₁ = 48 °C = 321.15 K

p₂ = 159 kPa × 1 atm/101.325 kPa = 1.569 atm

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T₂ = ?

=====

Calculation:

T₂ = 321.15 × 1.569/0.8882 × 2.0/0.718  

T₂ = 321.15 × 1.766 × 2.786  

T₂ = 321.15 × 1.569/0.8882 × 7.786  

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T₂ = 1900 °C

<em>Note</em>: The answer can have only <em>two</em> significant figures because that is all you gave for the second volume of the gas.

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