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3 years ago

In the determining molar volume experiment, we will use water. how will vapor pressure play a role in this experiment?

Chemistry

2 answers:

Aleonysh [2.5K]3 years ago

8 0

Answer:

The sample's volume is affected by the transferring molecules to the vapor phase, underestimating it.

Explanation:

Hello,

Molar properties are said to be measured per unit of mole of the involved substance; more specifically speaking, the molar volume accounts for the volume occupied per mole of the considered substance. In this case, since the affection of the vapor pressure is mean to be explained, we should assume that the experiment is carried out to determine a liquid's molar volume, in such a way, when performing the experiment one must remember that all the liquids have a vapor pressure associated with the pressure exerted by the in-equilibrium molecules of its vapor phase, besides, the vapor pressure causes the liquid to slowly boil as the time go by. In this manner, the vapor pressure could diminish the liquid's molar volume as it is progressively boiled causing an underestimation.

Best regards.

Bas_tet [7]3 years ago

7 0

Molar volume is a property of a component in a solution. It is defined as the volume occupied by one mole of the component in the closed system. You would not expect all solutions to execute volume additivity because intermolecular forces between the components come into play. There is no such thing as conservation of volume.

Vapor pressure affects molar volume because gases are very sensitive by these process conditions. Vapor pressure is very temperature-dependent. Consequently, at a different temperature, your component could expand or compress, thus, affecting the molar volume. Moreover, the pressure affects the molecular collisions in the system.

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Answer:

The answer to the questions are as follows

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(b) Reaction B is faster.

Explanation:

To solve the question we are meant to compare both reactions to see which one is faster

The values of the given activation energies are as follows

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Ea = 95.00 kJ mol–1 (22.71 kcal mol–1) and

for B

Ea = 74.20 kJ mol–1 (17.73 kcal mol–1)

T is the same for both reactions and is equal to 298 k

Concentration of both reaction = 1M

The Arrhenius Law is given by

k = $Ae^{\frac{-E_{a} }{RT} }$

Where

k = rate constant

Ea = activation energy

R = universal gas constant

T = temperature (Kelvin )

A = Arrhenius factor

Therefore

For reaction A, the rate constant k₁ is given by k₁ = $Ae^{\frac{-95000}{(8.314)(298)} }$

And for B the rate constant k₂ is given by k₂ = $Ae^{\frac{-74200 }{(8.314)(298)} }$

k₁ = A×2.225×10⁻¹⁷

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Answer:

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