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Harrizon [31]
3 years ago
7

C2H4(g) + H2(g) mc030-1.jpg C2H6(g) Which change would likely cause the greatest increase in the rate of the reaction?

Chemistry
2 answers:
amm18123 years ago
8 0
<h3><u>Answer;</u></h3>

increasing temperature and pressure

<h3><u>Explanation;</u></h3>
  • The reaction involves gases;

<em>C2H4(g) + H2(g) → C2H6(g)</em>

  • Therefore, <em><u>a change in pressure and temperature greatly affects the rate of reaction or the rate at which the products are formed.</u></em>
  • <em><u>Increasing the pressure in the reaction favors the rate of reaction</u></em> as the 2 volumes of the reactants forms 1 volume of the product and thus compression is evident, which will increase the rate of reaction.
  • <em><u>Increasing the temperature also increases the rate of reaction due to increases in the kinetic energy of molecules of the reactants and thus an increase in the number of collision by the particles of the reactants </u></em>and thus the speed at which the products are formed.
Vinil7 [7]3 years ago
5 0
The reaction is a hydrogenation reaction of an alkene, and its equation is:
C₂H₄(g) + H₂(g) → C₂H₆(g)
Therefore, this reaction can be sped up just as any other irreversible reaction may have its rate increased, by increasing temperature and pressure to increase the effective collisions of molecules.
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Which scientist is known for developing the planetary model of the atom
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Answer:

Ernest Rutherford

Explanation:

The Rutherford model, proposed originally in 1911, dipicts a planetary model of the atom in which there is a nucleus and electrons circling it.

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The reaction described by H2(g)+I2(g)⟶2HI(g) has an experimentally determined rate law of rate=k[H2][I2] Some proposed mechanism
MatroZZZ [7]

Answer:

Mechanism A and B are consistent with observed rate law

Mechanism A is consistent with the observation of J. H. Sullivan

Explanation:

In a mechanism of a reaction, the rate is determinated by the slow step of the mechanism.

In the proposed mechanisms:

Mechanism A

(1) H2(g)+I2(g)→2HI(g)(one-step reaction)

Mechanism B

(1) I2(g)⇄2I(g)(fast, equilibrium)

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Mechanism C

(1) I2(g) ⇄ 2I(g)(fast, equilibrium)

(2) I(g)+H2(g) ⇄ HI(g)+H(g) (slow)

(3) H(g)+I(g)→HI(g) (fast)

The rate laws are:

A: rate = k₁ [H2] [I2]

B: rate = k₂ [H2] [I]²

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]

<em>Where K' = K1 * K2</em>

C: rate = k₁ [H2] [I]

As:

K-1 [I]² = K1 [I2]:

rate = k' [H2] [I2]^1/2

Thus, just <em>mechanism A and B are consistent with observed rate law</em>

In the equilibrium of B, you can see the I-I bond is broken in a fast equilibrium (That means the rupture of the bond is not a determinating step in the reaction), but in mechanism A, the fast rupture of I-I bond could increase in a big way the rate of the reaction. Thus, just <em>mechanism A is consistent with the observation of J. H. Sullivan</em>

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