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prisoha [69]
3 years ago
5

If two reactions sum to an overall reaction, and the equilibrium constants for the two reactions are K1K1 and K2K2, what is the

equilibrium constant for the overall reaction?
Chemistry
1 answer:
inessss [21]3 years ago
7 0

Answer: The equilibrium constant for the overall reaction is K_1\times K_2

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.

a) A\rightarrow B+C

K_1=\frac{[B]\times [C]}{[A]}

b) B+C\rightarrow D

K_2=\frac{[D]}{[B]\times [C]}

For overall reaction on adding a and b we get c

c) A\rightarrow D

K_3=\frac{[D]}{[A]}

K_3=K_1\times K_2=\frac{[B]\times [C]}{[A]}\times \frac{[D]}{[B]\times [C]}

K_3=K_1\times K_2=\frac{[D]}{[A]}

The equilibrium constant for the overall reaction is K_1\times K_2

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Answer: The empirical formula for the given compound is NO_2

Explanation : Given,

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To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.370g}{16g/mole}=0.0231moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{0.130g}{14g/mole}=0.00928moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00928 moles.

For Oxygen  = \frac{0.0231}{0.00928}=2.4\approx 2

For Nitrogen = \frac{0.00928}{0.00928}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of O : N = 2 : 1

Hence, the empirical formula for the given compound is NO_2

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Please help me with this homework I really need help with this homework
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Answer:

For the first question its C, Gas

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