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WINSTONCH [101]
2 years ago
10

QUESTION 6 Consider the following reaction between the diatomic and monatomic forms of iodine: I2 (g) <-> 2I (g) When 0.09

5 M I2 is initially placed in a previously empty container and sealed, the system slowly reaches equilibrium. When equilibrium is reached, it is found that there is an equilibrium concentration of 0.0055 M of the monatomic form of iodine. Calculate the (unitless) equilibrium constant Kc. Round your answer to two sig figs, and express it in scientific notation.
Chemistry
1 answer:
aalyn [17]2 years ago
3 0

Answer: The equilibrium constant is 3.3\times 10^{-4}

Explanation:

Initial concentration of I_2 = 0.095 M  

The given balanced equilibrium reaction is,

                            I_2(g)\rightleftharpoons 2I(g)

Initial conc.          0.095 M       0 M

At eqm. conc.     (0.095-x) M   (2x) M

Given : 2x = 0.0055

x = 0.00275

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[l]^2}{[I_2]}  

Now put all the given values in this expression, we get :

K_c=\frac{(0.0055)^2}{(0.095-0.00275)}

K_c=\frac{(0.0055)^2}{0.09225}=0.00033

Thus the equilibrium constant is 3.3\times 10^{-4}

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If I had 3.50 x 10 24molecules of Cl2 gas, how many grams is this?
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Answer:

412 g Cl₂

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 3.50 × 10²⁴ molecules Cl₂

[Solve] grams Cl₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of Cl₂ - 2(35.45) = 70.9 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 3.50 \cdot 10^{24} \ molecules \ Cl_2(\frac{1 \ mol \ Cl_2}{6.022 \cdot 10^{23} \ molecules \ Cl_2})(\frac{70.9 \ g \ Cl_2}{1 \ mol \ Cl_2})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 412.072 \ g \ Cl_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

412.072 g Cl₂ ≈ 412 g Cl₂

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