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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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4 0
3 years ago
Calculate ∆G ◦ r for the decomposition of mercury(II) oxide 2 HgO(s) → 2 Hg(ℓ) + O2(g) ∆H◦ f −90.83 − − (kJ · mol−1 ) ∆S ◦ m 70.
bagirrra123 [75]

Answer:

4. +117,1 kJ/mol

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ΔG of a reaction is:

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For the reaction:

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ΔHr: 2ΔHf Hg(l) + ΔHf O₂(g) - 2ΔHf HgO(s)

As ΔHf of Hg(l) and ΔHf O₂(g) are 0:

ΔHr: - 2ΔHf HgO(s) = <u><em>181,66 kJ/mol</em></u>

<u><em /></u>

In the same way ΔSr is:

ΔSr= 2ΔS° Hg(l) + ΔS° O₂(g) - 2ΔS° HgO(s)

ΔSr= 2* 76,02J/Kmol + 205,14 J/Kmol - 2*70,19 J/Kmol

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Thus, ΔGr at 298K is:

ΔGr = 181,66 kJ/mol - 298K*0,216kJ/Kmol

ΔGr = +117,3 kJ/mol ≈ <em>4. +117,1 kJ/mol</em>

<em></em>

I hope it helps!

5 0
3 years ago
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Answer:

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