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3 years ago

Given the value of the equilibrium constant (Kc) for the equation (a), calculate the equilibrium constant for equation (b)

Chemistry

1 answer:

matrenka [14]3 years ago

5 0

Answer: The value of equilibrium constant for new reaction is $1.92\times 10^{-25}$

Explanation:

The given chemical equation follows:

$O_2(g)\rightarrow \frac{2}{3}O_3(g)+\frac{1}{2}O_2(g)$

The equilibrium constant for the above equation is $5.77\times 10^{-9}$

We need to calculate the equilibrium constant for the equation of 3 times of the above chemical equation, which is:

$3O_2(g)\rightarrow 2O_3(g)$

The equilibrium constant for this reaction will be the cube of the initial reaction.

If the equation is multiplied by a factor of '3', the equilibrium constant of the new reaction will be the cube of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{eq}'=(5.77\times 10^{-9})^3=1.92\times 10^{-25}$

Hence, the value of equilibrium constant for new reaction is $1.92\times 10^{-25}$

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Aluminum costs about 85 cents per pound (453.592 g). How much is one atom of aluminum?

kakasveta [241]

One aluminum atom will cost 0.84 × 10⁻²³ cents.

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To calculate the number of aluminum atoms we use the Avogadro's number:

if 1 mole of aluminum contains 6.022 × 10²³ aluminum atoms

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if 101.17 × 10²³ aluminum atoms costs 85 cents

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Y = (1 × 85) / (101.17 × 10²³) = 0.84 × 10⁻²³ cents

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Avogadro's number

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