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4 years ago

If a balloon is filled with a mixture of helium and oxygen, which gas will escape faster? Why?

Chemistry

1 answer:

Serjik [45]4 years ago

4 0

Answer:

The rate of leakage will be higher for helium; its molecules move about 3 times faster than oxygen’s

Explanation:

Step 1: Data given

Molar mass helium = 4.0 g/mol

Molar mass O2 = 32 g/mol

Step 2: Graham's law

Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of the molecular mass : 1/(Mr)^0.5

Rate of escape for He = 1/(4.0)^0.5 = 0.5

Rate of escape for O2 = 1/(32)^0.5 = 0.177

The rate of leakage will be higher for helium; its molecules move about 3 times faster than oxygen’s

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The problem=the equation

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3 years ago

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting e

sp2606 [1]

The question is incomplete, here is the complete question:

While ethanol is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene with water vapor at elevated temperatures.

A chemical engineer studying this reaction fills a 1.5 L flask at 12°C with 1.8 atm of ethylene gas and 4.7 atm of water vapor. When the mixture has come to equilibrium she determines that it contains 1.16 atm of ethylene gas and 4.06 atm of water vapor.

The engineer then adds another 1.2 atm of ethylene, and allows the mixture to come to equilibrium again. Calculate the pressure of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

Answer: The partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

Explanation:

We are given:

Initial partial pressure of ethylene gas = 1.8 atm

Initial partial pressure of water vapor = 4.7 atm

Equilibrium partial pressure of ethylene gas = 1.16 atm

Equilibrium partial pressure of water vapor = 4.06 atm

The chemical equation for the reaction of ethylene gas and water vapor follows:

$CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)$

Initial: 1.8 4.7

At eqllm: 1.8-x 4.7-x

Evaluating the value of 'x'

$\Rightarrow (1.8-x)=1.16\\\\x=0.64$

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{CH_3CH_2OH}}{p_{CH_2CH_2}\times p_{H_2O}}$

$p_{CH_2CH_2}=1.16atm\\p_{H_2O}=4.06atm\\p_{CH_3CH_2OH}=0.64atm$

Putting values in above expression, we get:

$K_p=\frac{0.64}{1.16\times 4.06}\\\\K_p=0.136$

When more ethylene is added, the equilibrium gets re-established.

Partial pressure of ethylene added = 1.2 atm

$CH_2CH_2(g)+H_2O(g)\rightleftharpoons CH_3CH_2OH(g)$

Initial: 2.36 4.06 0.64

At eqllm: 2.36-x 4.06-x 0.64+x

Putting value in the equilibrium constant expression, we get:

$0.136=\frac{(0.64+x)}{(2.36-x)\times (4.06-x)}\\\\x=0.363,13.41$

Neglecting the value of x = 13.41 because equilibrium partial pressure of ethylene and water vapor will become negative, which is not possible.

So, equilibrium partial pressure of ethanol = (0.64 + x) = (0.64 + 0.363) = 1.003 atm

Hence, the partial pressure of ethanol after equilibrium is reached the second time is 1.0 atm

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The SI unit will be

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