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Arte-miy333 [17]
3 years ago
14

What may be expected when K < 1.0?

Chemistry
2 answers:
Romashka [77]3 years ago
8 0
The reaction will generally form more reactants than products.
IceJOKER [234]3 years ago
8 0

Answer: The concentration of one or more of the products is small.

The reaction will not proceed very far to the right.

The reaction will generally form more reactants than products.

Explanation:

Equilibrium constant is defined as the ratio of concentration of product to the concentration of reactants each raised to the power their stoichiometric ratios. It is represented by the symbol 'K'. For the general equilibrium equation:

aA+bB\rightleftharpoons cC+dD

The expression for equilibrium constant is given as:

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

There are three cases:  

When K > 1; the reaction is product favored.

When K = 1; the reaction is at equilibrium.

When K < 1; the reaction is reactants favored i.e will shift to left and thus concentration of product is less as compared to concentration of reactant.

Thus, the correct statements are

The concentration of one or more of the products is small.

The reaction will not proceed very far to the right.

The reaction will generally form more reactants than products.

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Equal molar quantities of Ca2 and EDTA (H4Y) are added to make a 0.010 M solution of CaY2- at pH 10. The formation constant for
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the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

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Given the data in the question;

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Kf = CaY^{2- / ( [Ca^{2+][y^{4-] ) =  CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ CaY^{2- / ( [Ca^{2+][\alpha _4CH_4Y] )  = 5.0 × 10¹⁰

⇒ 0.010 / ( [x][ 0.35 × x] )  = 5.0 × 10¹⁰

⇒ 0.010 / 0.35x²  = 5.0 × 10¹⁰

⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 5.7142857 × 10⁻¹³

⇒ x = √(5.7142857 × 10⁻¹³)

⇒ x = 7.559 × 10⁻⁷

Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

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