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snow_lady [41]
3 years ago
9

Consider this equilibrium reaction at 400 K. Br2(g)+Cl2(g)↽−−⇀2BrCl(g)Kc=7.0 If the composition of the reaction mixture at 400 K

is [BrCl]=0.00415 M, [Br2]=0.00366 M, and [Cl2]=0.000672 M, what is the reaction quotient, ???? ? ????= How is the reaction quotient related to the equilibrium constant, Kc, for this reaction?
Chemistry
1 answer:
Genrish500 [490]3 years ago
4 0

Answer:

Q = 7.0

Q = kc. The reaction is in equilibrium

Explanation:

Based on the reaction:

Br₂ + Cl₂ ⇄ 2BrCl

Equilibrium constant of the reaction, kc, is the ratio of <em>equilibrium concentrations</em> products over reactants powered to its reaction coefficient:

Kc = [BrCl]² / [Br₂] [Cl₂] = 7.0

Now, reaction quotient, Q, is write as the same Kc but the concentrations are actual concentrations:

Q = [BrCl]² / [Br₂] [Cl₂]

Replacing:

Q = [0.00415M]² / [0.00366M] [0.000672M]

Q = 7.0

Now, as Q = Kc = 7.0, the reaction mixture is in equilibrium

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<u>Answer:</u> S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

<u>Explanation:</u>

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when oxidation number of a species decreases.

For the given chemical reaction:

HNO_3+S\rightarrow H_2SO_4+NO

<u>On the reactant side:</u>

Oxidation number of H = +1

Oxidation number of N = +5

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Oxidation number of S = 0

<u>On the product side:</u>

Oxidation number of H = +1

Oxidation number of N = +2

Oxidation number of O = -2

Oxidation number of S = +6

As the oxidation number of S is increasing from 0 to +6. Thus, it is getting oxidized. Similarly, the oxidation number of N is decreasing from +5 to +2. Thus, it is getting reduced.

The oxidation numbers of O and H remain the same on both sides of the reaction. Thus, they are neither getting oxidized or reduced.

Hence, S is getting oxidized, N is getting reduced and O and H undergo no oxidation or reduction

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3 years ago
Suppose you have a mixture of solid mg(oh)2 and baso4, how do you separate the mixture
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The aqueous solution will now contain MgSO₄ and BaSO₄. Unlike BaSO₄, MgSO₄ is soluble in water. So, you filter out the solution. You can set aside the BaSO₄ on the filter paper. To retrieve Mg(OH)₂, add NaOH.

MgSO₄ + 2 NaOH = Mg(OH)₂ + Na₂SO₄

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Explanation:

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Answer : The concentration of HI and I_2 at equilibrium is, 0.0158 M and 0.00302 M respectively.

Explanation :

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Now we have to calculate the value of equilibrium constant (K).

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Initial conc.      0.0163     0.00415      0.00276

At eqm.        (0.0163-2x) (0.00415+x)  (0.00276+x)

As we are given:

Concentration of H_2 at equilibrium = 0.00467 M

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(0.00415+x) = 0.00467

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Concentration of I_2 at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M

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Answer:

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