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kakasveta [241]
3 years ago
7

Is a forward or reverse reaction favored?

Chemistry
1 answer:
Alex Ar [27]3 years ago
6 0

Answer:

(a) CH_3COOH \rightleftharpoons CH_3COO^-+ H^+: reverse reaction is favored.

(b) AgCl \rightleftharpoons Ag^+ + Cl^-: reverse reaction is favored.

(c) Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^-: reverse reaction is favored.

(d) A+B\rightleftharpoons C: forward reaction is favored.

Explanation:

Hello,

(a) CH_3COOH \rightleftharpoons CH_3COO^-+ H^+:

In this case, since the Ka is lower than 1, we infer the reverse reaction is favored since the reactant (acetic acid) will tend to have a higher concentration.

(b) AgCl \rightleftharpoons Ag^+ + Cl^-:

In this case, since the Ka is lower than 1, we infer the reverse reaction is favored since the reactant (silver chloride) will tend to have a higher concentration.

(c) Al(OH)_3 \rightleftharpoons Al^{3+} + 3OH^-

In this case, since the Ka is lower than 1, we infer the reverse reaction is favored since the reactant (aluminium hydroxide) will tend to have a higher concentration.

(d) A+B\rightleftharpoons C

In this case, since the Ka is greater than 1, we infer the forward reaction is favored since the product (C) will tend to have a higher concentration.

Regards.

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Answer:

Ke = 34570.707

Explanation:

  • H2(g) + Br2(g) → 2 HBr(g)

equilibrium constant (Ke):

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∴ [HBr] = (37.0 mol) / (2 L) = 18.5 mol/L

∴ [Br2] = (0.110 mol) / (2 L) = 0.055 mol/L

∴ [H2] = (0.360 mol) / (2 L) = 0.18 mol/L

⇒ Ke = (18.5 mol/L)² / (0.055 mol/L)(0.18 mol/L)

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