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Vlad1618 [11]
3 years ago
12

At equilibrium, the concentrations of reactants and products can be predicted using the equilibrium constant, Kc, which is a mat

hematical expression based on the chemical equation. For example, in the reaction aA+bB⇌cC+dD where a, b, c, and d are the stoichiometric coefficients, the equilibrium constant is Kc=[C]c[D]d[A]a[B]b where [A], [B], [C], and [D] are the equilibrium concentrations. If the reaction is not at equilibrium, the quantity can still be calculated, but it is called the reaction quotient, Qc, instead of the equilibrium constant, Kc. Qc=[C]tc[D]td[A]ta[B]tb where each concentration is measured at some arbitrary time t. Part A A mixture initially contains A, B, and C in the following concentrations: [A] = 0.300 M , [B] = 1.10 M , and [C] = 0.450 M . The following reaction occurs and equilibrium is established: A+2B⇌C At equilibrium, [A] = 0.110 M and [C] = 0.640 M . Calculate the value of the equilibrium constant, Kc. Express your answer numerically.
Chemistry
1 answer:
mr Goodwill [35]3 years ago
8 0

Answer:

The equillibrium constant Kc = 11.2233

Explanation:

Step 1:

aA + bB ⇔ cC + dD

with a, b, c and d = coefficients

Kc = equillibrium constant =( [C]^c [D]^d ) / ( [A]^a [B]^b)

Concentration at time t

[A] = 0.300 M

[B] =1.10 M

[C] = 0.450 M

 

Change :

A: -x

B: -2x

C: -x

The following reaction occurs and equillibrium is established

A + 2B ⇔ C

[A] = 0.110M

[B] = ?

[C] = 0.640 M

For A we see that after change: 0.3 -x = 0.11

Then for B we have  1.1 - 2x = ? ⇒ 1.1 -2 *0.19 = 0.72

This gives us for the equillibrium constant Kc = [C] / [A][B] ²

Kc = 0.64 / (0.11) * (0.72)² = 11.2233

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