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Margarita [4]
3 years ago
12

What can be said about an endothermic reaction with a negative entropy change? spontaneous at all temperatures. spontaneous at h

igh temperatures. spontaneous at low temperatures. spontaneous in the reverse direction at all temperatures. nonspontaneous in either direction at all temperatures. Submit
Chemistry
1 answer:
Arisa [49]3 years ago
4 0

Answer:

  • <em>spontaneous in the reverse direction at all temperatures</em>

Explanation:

The parameter to predict the spontaneity of a reduction is the change in the free energy, ΔG.

This is how it works:

  • If ΔG = 0, the system is in equilibrium
  • If ΔG > 0, the reaction is non-spontaneous.
  • If ΔG < 0, the reaction is spontaneous.

In brief, free energy's decrease tells that a reaction is spontaneous, while free energy's increase tells a reaction is nonspontaneous.

Now, take the definition of the free energy:

  • G = H - TS
  • ΔG = ΔH - TΔS

The conditions given in the statement are:

  • <em>An endothermic reaction</em> ⇒ ΔH > 0 (positive)
  • <em>A negative entropy change</em> ⇒ ΔS < 0 ⇒ TΔS < 0 (negative)

Replacing in the equation ΔG = ΔH - TΔS, you get:

  • ΔG = positive - (negative) = positive + positive = positive.

Then, you conclude that <em>for an endothermic reaction with a negative entropy change, </em>the change in the free energy is positive, and so the reaction is nonspontaneous (at all temperatures) in the forward direction.

Since, the change in the reverse direction has opposite sign, you also conclude that <em><u>the reaction is spontaneous in the reverse direction at all temperatures.</u></em>

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Answer:

(a) See below

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Explanation:

The equation relating the temperature to time is

T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )

1. Calculate the thermometer readings after  0.5 min and 1 min

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\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}

(b) After 1 min

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

  t = t - 1, because the cooling starts 1 min late

\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}

3. Plot the temperature readings as a function of time.

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Explanation:

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The properties of the mixture like concentration may change for different parts of the mixture.

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