Mastering Chemistry Equilibrium Constant and Reaction Quotient Part A At equilib

Question

Mastering Chemistry

Equilibrium Constant and Reaction Quotient Part A At equilibrium, the concentrations of reactants and A mixture initially contains A, B. and C in the following concentrations: A-0.600 MB-1.35 M and [C] = 0.700 M . The following reaction occurs and equilibrium is established produ constant, Kc, which is a mathematical expression based on the chemical equation. For example, in the reaction roducts can be predicted using the equilibrium Atequilibrium, [A] = 0470 M and [C] = 0.830 M. Calculate the value of the equilibrium constant, Express your answer numerically where a, b, c, and d are the stoichiometric coefficients, the equilibrium constant is where [A][B], [CI, 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, c Submit Hints My Answers Give Up Review Part Incorrect Try Again, 9 attempts remaining At B where each concentration is measured at some arbitrary time t Part B

Explanation / Answer

A)

consider the given reaction

A + 2B ---> C

initially

[A] = 0.6

[B] = 1.35

[C] = 0.7

now at equilibrium

[A]eq = 0.6 - x

[B]eq = 1.35 -2x

[C]eq = 0.7 + x

given

[A]eq = 0..47

so

0.6 - x = 0.47

x = 0.13

so

[B]eq = 1.35 - ( 2 * 0.13) = 1.09

[C]eq = 0.7 + 0.13 = 0.83

now

Kc = [C] / [A] [B]^2

Kc = [0.83] / [0.47] [1.09]^2

Kc = 1.4864

2)

we know that

concentration = moles / volume (L)

so

[C] = 9.5 / 3.75 = 2.5333

[H20] = 12.2 / 3.75 = 3.25333

[CO] = 3.2 / 3.75= 0.85333

[H2] = 6.8 / 3.75 = 1.81333

now

consider the given reaction

C (s) + H20 --> CO + H2

the reaction quotient is

Q = [CO] [H2] / [H20]

Q = [0.85333] [1.81333] / [3.25333]

Q = 0.4756

3) Part A

the reaction is

H2 + I2 --> 2HI

the value of Kc is given by

Kc = [HI]^2 / [H2] [I2]

Kc = [0.922]^2 / [0.0355] [0.0388]

Kc = 617.16

4) Part B:

Kc = [HI]^2 / [H2] [I2]

9.6 = [0.387]^2 / [H2] [4.6 x 10-2]

[H2] = 0.339 M

5) Part C :

Kc = [HI]^2 /[H2] [I2]

50.2 = [HI]^2 / [4.8 x 10-2] [4.72 x 10-2]

[HI] = 0.337 M

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