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-BARSIC- [3]
3 years ago
11

For the reaction show below determine the rate law. H2SO3(aq) + 6I-(aq) + 4H+(aq) → S(s) + 2I3- + 3H2O (l) In the presence of a

little starch, this solution turns from colorless to blue when I3- is formed. The following data were obtained: [H2SO3] [H+] [I-] Initial Rate (mol/L s) 1.0 x 10-2 2.0 x 10-1 2.0 x 10-1 1.66 x 10-7 3.0 x 10-2 2.0 x 10-1 2.0 x 10-1 4.98 x 10-7 1.0 x 10-2 1.0 x 10-1 2.0 x 10-1 8.30 x 10-8 1.0 x 10-2 2.0 x 10-1 4.0 x 10-1 13.2 x 10-7 Calculate the rate law for this reaction: rate = k [H2SO3] [H+] [I-

Chemistry
1 answer:
alekssr [168]3 years ago
7 0

Answer:

Please see the attached picture for the complete answer.

Explanation:

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A) The catalyzed reaction passes through C.

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In which of the following substances does sharing of electrons between two atoms occur?
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  • PO4∧-3

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A closed system initially containing 1×10^-3 hydrogen 2×10^-3M iodine at 448 degree Celsius and is allowed to reach equilibrium.
GaryK [48]

Answer:

Kc = 50.5

Explanation:

We determine the reaction:

H₂  +  I₂   ⇄   2HI

Initially we have 0.001 molesof H₂

and 0.002 moles of I₂

If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.

           H₂     +      I₂      ⇄   2HI

In:     0.001       0.002           -

R:       x                 x                2x

Eq:  0.001-x    0.002-x      0.00187  

x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted

So in the equilibrium we have:

0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵  moles of H₂

0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂

Expression for Kc is =  (HI)² / (H₂) . (I₂)

0.00187 ² /  6.5×10⁻⁵ . 1.065×10⁻³ = 50.5

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