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podryga [215]
3 years ago
14

Which of the following types of reactions would decrease the entropy within a cell? A. Dehydration reaction, B. Hydrolysis, C. R

espiration, D. Digestion, E. Catabolism
Chemistry
1 answer:
kicyunya [14]3 years ago
6 0
<h3><u>Answer;</u></h3>

A. Dehydration reactions.

<h3><u>Explanation</u>;</h3>
  • Dehydration reactions refer to chemical events where a water molecule is lost because of the presence of another reacting molecule.
  • Dehydration reaction is said to cause a decrease in entropy in various cells.
  • <em><u>Entropy within a cell refers to a state of inner instability secondary to different energies provided by the different molecular components.</u></em>
  • <em><u>When the water component is removed,  there will be less molecular energy and therefore decreased entropy.</u></em>
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Suppose a 250. mL flask is filled with 0.30 mol of N_2 and 0.70 mol of NO. The following reaction becomes possible:N_2(g) +O2 →
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0.4 M

Explanation:

Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.

Because there's no O₂ in the beginning, the NO will decompose:

N₂(g) + O₂(g) ⇄ 2NO(g)

0.30 0 0.70 Initial

+x +x -2x Reacts (the stoichiometry is 1:1:2)

0.30+x x 0.70-2x Equilibrium

The equilibrium concentrations are the number of moles divided by the volume (0.250 L):

[N₂] = (0.30 + x)/0.250

[O₂] = x/0.25

[NO] = (0.70 - 2x)/0.250

K = [NO]²/([N₂]*[O₂])

K = \frac{(\frac{0.70 -2x}{0.250})^2 }{\frac{0.30+x}{0.250}*\frac{x}{0.250} }

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7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)

4x² - 2.80x + 0.49 = 2.31x + 7.70x²

3.7x² + 5.11x - 0.49 = 0

Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70

x = 0.09 mol

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[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M

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The energy transforms from chemical energy to heat and light energy. Because when the candle burns a chemical reaction occurs, and produces heat and light.
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