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pychu [463]
4 years ago
9

The equilibrium-constant expression for a reaction written in one direction is the __________ of the one for the reaction writte

n for the reverse direction.
Chemistry
1 answer:
VLD [36.1K]4 years ago
8 0

Answer:

  • <u>    reciprocal   </u>

Explanation:

The <em>equilibrium constant</em> for an <em>equilibrium reaction </em>is the ratio of the equilibrium constant for the forward reaction, Kf, to the equilibrium constant for the reverse reaction, Kr:

  • For A ⇄ B

  • The forward reaction is: A → B, with rate constant is Kf = K₁ and
  • The reverse reaction is: B → A, with rate constant Kr = K₂

            K_{eq}=\dfrac{K_f}{K_r}=\dfrac{K_1}{K_2}

When you write the reaction in the other reaction, the forward and the reverse reaction are exchanged:

  • For B ⇄ A

  • The forward reaction is B → A, with rate constant Kf = K₂
  • The reverse reaction is A → B, with rate constant Kr = K₁

             

             K'_{eq}=\dfrac{K_f}{K_r}=\dfrac{K_2}{K_1}

As you see:

          K'_{eq}=\dfrac{1}{K_{eq}}

Thus, <em>the equilibrium-constant expression for a reaction written in one direction is the </em><em><u>    reciprocal</u></em><em>___</em><em> of the one for the reaction written for the reverse direction.</em>

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Match the term with the definition.
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A lab technician needs to create 570.0 milliliters of a 2.00 M solution of magnesium chloride (MgCl2). To make this solution, ho
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Answer:

108.3g

Explanation:

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Molarity of solution = 2M

Unknown

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Solution

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Using this known moles, we can the unknown mass of MgCl₂ the technician would require:

       Mass of MgCl₂ required = number of moles of MgCl₂ x molar mass

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7 0
3 years ago
A student is doing an experiment to determine the effects of temperature on an object. He writes down that the initial temperatu
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Answer:

1) The Kelvin temperature cannot be negative

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- Kelvin scale: the Kelvin is indicated with K. it is based on the concept of "absolute zero" temperature, which is the temperature at which matter stops moving, and it is placed at zero Kelvin (0 K), so this scale cannot have negative temperatures, since 0 K is the lowest possible temperature.

The expression to convert from Celsius degrees to Kelvin is:

T(K)=T(^{\circ}C)+273.15

Therefore  in this problem, since the student reported a temperature of -3.5 ºK, the errors done are:

1) The Kelvin temperature cannot be negative

2) The Kelvin degree is written as K, not ºK

6 0
4 years ago
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