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3 years ago

0.7 L of H2 enters a reaction at standard temperature and pressure.

Chemistry

1 answer:

ASHA 777 [7]3 years ago

6 0

Answer:

1.88 × 10²³ particles

Explanation:

Given data:

Volume of H₂ = 0.7 L

Number of particles at STP = ?

Solution:

First of all we will calculate the number of moles of H₂.

PV = nRT

n = PV / RT

n = 1 atm . 7 L / 0.0821 L. atm. mol⁻¹. k⁻¹ × 273.15 K

n = 7 atm. L / 22.43 L. atm. mol⁻¹

n = 0.312 mol

it is known that,

2 g H₂ = 1 mol = 6.022 × 10²³ particles

For 0.312 mol

0.312 × 6.022 × 10²³ particles

1.88 × 10²³ particles

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<u>Answer:</u> The concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.

<u>Explanation:</u>

For the given chemical reaction:

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

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Concentration of hydrogen gas when equilibrium is re-established = 0.1 + 0.15 = 0.25 M

Now, the equilibrium is shifting to the reactant side. The equation follows:

$CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)$

Initial: 0.1 0.1 0.4 0.1

At eqllm: 0.1+x 0.1+x 0.4-x 0.25-x

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Neglecting the negative value of 'x'

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Concentration of hydrogen gas = (0.25 - x) = (0.25 - 0.038) = 0.212 M

Concentration of carbon monoxide = (0.1 + x) = (0.1 + 0.038) = 0.138 M

Concentration of water = (0.1 + x) = (0.1 + 0.038) = 0.138 M

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