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Alex
3 years ago
6

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 and respectively. The value of Keq for this reaction is__________.
Chemistry
1 answer:
bearhunter [10]3 years ago
4 0

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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