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3 years ago

SCICLUL CUCCIUCUCURUCUS

Chemistry

1 answer:

Evgesh-ka [11]3 years ago

7 0

Answer:

hi there!

the correct answer is atomic number 87, francium.

Explanation:

elements become more reactive as you go from the top left starting with hydrogen down to the bottom right

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3 years ago

Consider the following chemical reaction: H2 (g) + I2 (g) 2HI (g) At equilibrium in a particular experiment, the concentrations

bearhunter [10]

The question is incomplete, here is the complete question:

Consider the following chemical reaction: H₂ (g) + I₂ (g) ⇔ 2HI (g) At equilibrium in a particular experiment, the concentrations of H₂, I₂, and HI were 0.15 M, 0.033 M and 0.55 M respectively. The value of Keq for this reaction is

<u>Answer:</u> The value of $K_{eq}$ for the given reaction is 61.11

<u>Explanation:</u>

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_[eq}$

For a general chemical reaction:

$aA+bB\rightarrow cC+dD$

The expression for $K_{eq}$ is written as:

$K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

For the given chemical equation:

$H_2(g)+I_2(g)\rightleftharpoons 2HI(g)$

The expression of $K_{eq}$ for above equation follows:

$K_{eq}=\frac{[HI]^2}{[H_2][I_2]}$

We are given:

$[HI]_{eq}=0.55M$

$[H_2]_{eq}=0.15M$

$[I_2]_{eq}=0.033M$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.55)^2}{0.15\times 0.033}\\\\K_{eq}=61.11$

Hence, the value of $K_{eq}$ for the given reaction is 61.11

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