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vivado [14]
3 years ago
6

Which represents the correct equilibrium constant expression for the reaction below?

Chemistry
2 answers:
charle [14.2K]3 years ago
6 0

Answer:

the answer is D

Explanation:3

ON EDG

Ad libitum [116K]3 years ago
4 0

Answer:

       \large\boxed{\large\boxed{K_c=\dfrac{[Cu^{2+}]}{[Ag^+]^2}}}

Explanation:

<u>1. Chemical equilibrium equation:</u>

    Cu(s)+2Ag^+(aq)\rightleftharpoons Cu^{2+}+2Ag(s)

<u>2. Species</u>

In an equilibrium constant expression you do not include the solid substances; only gases and dissolved substances.

The symbol (s) means solid, thus Cu(s) and Ag(s) shall not appear in your equilibrium constant expression.

The symbol (aq) means in aqueous solution, thus the both Ag^+ and Cu^{2+} must appear in the equilibrium constant expression.

<u>3. Equilibrium constant expression.</u>

It is the quotient of the product of the concentrations of the species on the right hand side of the equilibrium equation, each raised to its corresponding coefficient, and the product of the concentrations of the species on the left hand side, each raised to its coresponding coefficient.

        K_c=\dfrac{[Cu^{2+}]}{[Ag^+]^2}

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In the first 15.0 s of the reaction, 1.9×10−2 mol of O2 is produced in a reaction vessel with a volume of 0.480 L . What is the
N76 [4]

The question is incomplete, here is the complete question:

Consider the following reaction:  2N_2O(g)\rightarrow 2N_2(g)+O_2(g)

In the first 15.0 s of the reaction, 1.9×10⁻² mol of O₂ is produced in a reaction vessel with a volume of 0.480 L . What is the average rate of the reaction over this time interval?

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We are given:

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C_1 = initial concentration of oxygen gas = 0 M

t_2 = final time = 15.0 s

t_1 = initial time = 0 s

Putting values in above equation, we get:

\text{Average rate of appearance of }O_2=\frac{0.0396-0}{15-0}\\\\\text{Average rate of appearance of }O_2=2.64\times 10^{-3}M/s

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