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erastovalidia [21]
2 years ago
6

A study of the decomposition reaction 3RS2  3R + 6S yields the following initial rate dat

Chemistry
1 answer:
shutvik [7]2 years ago
3 0
Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150                0.0394
0.250                0.109
0.350                0.214
0.500                0.438</span>
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.
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Answer:

- 1273.02 kJ.

Explanation:

This problem can be solved using Hess's Law.

Hess's Law states that <em>regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function.</em>

  • We should modify the given 3 equations to obtain the proposed reaction:

<em>6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s),</em>

<em></em>

  • We should multiply the first equation by (6) and also multiply its ΔH by (6):

6C(s) + 6O₂(g) → 6CO₂(g), ∆H₁ = (6)(–393.51 kJ) = - 2361.06 kJ,

  • Also, we should multiply the second equation and its ΔH by (6):

6H₂(g) + 3O₂(g) → 6H₂O(l), ∆H₂ = (6)(–285.83 kJ) = - 1714.98 kJ.

  • Finally, we should reverse the first equation and multiply its ΔH by (- 1):

6CO₂(g) + H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g), ∆H₃ = (-1)(–2803.02 kJ) = 2803.02 kJ.

  • By summing the three equations, we cam get the proposed reaction:

<em>6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s),</em>

<em></em>

  • And to get the heat of reaction for the production of glucose, we can sum the values of the three ∆H:

<em>∆Hrxn = ∆H₁ + ∆H₂ + ∆H₃ =</em> (- 2361.06 kJ) + (- 1714.98 kJ) + (2803.02 kJ) = <em>- 1273.02 kJ.</em>

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