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kvasek [131]
3 years ago
13

If an equal number of moles of reactants are used, do the following equilibrium mixtures contain primarily reactants or products

? Please note this question has 1 submission for each part. (a) HCN(aq) + H2O(l) equilibrium reaction arrow CN −(aq) + H3O+(aq) K = 6.2 10-10 products reactants (b) H2(g) + Cl2(g) equilibrium reaction arrow 2 HCl(g) K = 2.51 104 reactants products
Chemistry
1 answer:
puteri [66]3 years ago
4 0

<u>Answer:</u>

<u>For a:</u> The equilibrium mixture contains primarily reactants.

<u>For b:</u> The equilibrium mixture contains primarily products.

<u>Explanation:</u>

There are 3 conditions:

  • When K_{eq}>1; the reaction is product favored.
  • When K_{eq}; the reaction is reactant favored.
  • When K_{e}=1; the reaction is in equilibrium.

For the given chemical reactions:

  • <u>For a:</u>

The chemical equation follows:

HCN(aq.)+H_2O(l)\rightleftharpoons CN^-(aq.)+H_3O^+(aq.);K_{eq}=6.2\times 10^{-10}

The expression of K_{eq} for above reaction follows:

K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}

As, K_{eq}, the reaction will be favored on the reactant side.

Hence, the equilibrium mixture contains primarily reactants.

  • <u>For b:</u>

The chemical equation follows:

H_2(g)+Cl_2(g)\rightleftharpoons 2HCl(g);K_{eq}=2.51\times 10^{4}

The expression of K_{eq} for above reaction follows:

K_{eq}=\frac{[HCl]^2}{[H_2][Cl_2]}=2.51\times 10^{4}

As, K_{eq}>1, the reaction will be favored on the product side.

Hence, the equilibrium mixture contains primarily products.

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