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At 2730 °C, hydrogen molecules dissociate into hydrogen atoms according to the equation H2 (g) → 2 H(g). 10.0 g H2 (g) is placed into a 100.0-L container, sealed and heated to 2730 °C so that the hydrogen molecules begin to dissociate. What is the partial pressure of hydrogen molecules and the total pressure within the container once 25% of H2 has dissociated?

Respuesta :

Answer:

Partial pressure of hydrogen molecule = 9·246 atm

Total pressure within the container = 15·41 atm

Explanation:

Given chemical equation

H2 (g) → 2 H(g)

For every 1 mole of hydrogen molecule dissociated, 2 moles of hydrogen atoms is formed

Given number of moles of hydrogen molecule = 10 ÷ 2 = 5

As 25% of H2 is dissociated, it means that 5 ÷ 4 moles of hydrogen molecule is dissociated

Number of moles of hydrogen atom formed = 2 × 1·25 = 2·5

Remaining moles of hydrogen molecule = 5 - 1·25 = 3·75

Assuming that both hydrogen molecule and hydrogen atom as ideal gases

Ideal gas equation = P × V = n × R × T

where P is the pressure of the gas

V is the volume occupied by the gas

n is the number of moles of gas

R is the ideal gas constant

T is the temperature of the gas

Partial pressure is defined as the pressure exerted by the gas when it occupies the complete volume of the container

Let the partial pressure of hydrogen molecule be P atm

Applying ideal gas equation to the hydrogen molecule

V = 100 L

n = 3·75

T = 2730°C = 2730 + 273 K = 3003 K

P × 100 = 3·75 × 0·0821 × 3003

∴ P = 9·246 atm

Let [tex]P_{1}[/tex] be the partial pressure of hydrogen atom

Applying ideal gas equation to the hydrogen atom

V = 100 L

n = 2·5

T = 2730°C = 2730 + 273 K = 3003 K

[tex]P_{1}[/tex] × 100 = 2·5 × 0·0821 × 3003

∴ [tex]P_{1}[/tex] = 6·164 atm

Total pressure within the container = P +  [tex]P_{1}[/tex] = 9·246 + 6·164 atm = 15·41 atm

∴ Total pressure = 15·41 atm

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