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8_murik_8 [283]
3 years ago
13

What can be said about an Endothermic reaction with a negative entropy change?The reaction isa. spontaneous at all temperatures.

b. spontaneous at high temperatures.c. spontaneous at low temperatures.d. spontaneous in the reverse direction at all temperatures.e. nonspontaneous in either direction at all temperatures.What can be said about an Exothermic reaction with a negative entropy change?The reaction isa. spontaneous at all temperatures.b. spontaneous at high temperatures.c. spontaneous at low temperatures.d. spontaneous in the reverse direction at all temperatures.e. nonspontaneous in either direction at all temperatures.
Chemistry
1 answer:
Lyrx [107]3 years ago
3 0

Answer:

1. d. The reaction is spontaneous in the reverse direction at all temperatures.

2. c. The reaction is spontaneous at low temperatures.

Explanation:

The spontaneity of a reaction is associated with the Gibbs free energy (ΔG). When ΔG < 0, the reaction is spontaneous. When ΔG > 0, the reaction is non-spontaneous. ΔG is related to the enthalpy (ΔH) and the entropy (ΔS) through the following expression:

ΔG = ΔH - T. ΔS  [1]

where,

T is the absolute temperature (T is always positive)

<em>1. What can be said about an Endothermic reaction with a negative entropy change?</em>

If the reaction is endothermic, ΔH > 0. Let's consider ΔS < 0. According to eq. [1], ΔG is always positive. The reaction is not spontaneous in the forward direction at any temperature. This means that the reaction is spontaneous in the reverse direction at all temperatures.

<em>2. What can be said about an Exothermic reaction with a negative entropy change?</em>

If the reaction is exothermic, ΔH < 0. Let's consider ΔS < 0. According to eq. [1], ΔG will be negative when |ΔH| > |T.ΔS|, that is, at low temperatures.

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Explanation:

The given data for case (1) is as follows.

  h = 20 cm = 0.2 m

Assuming that a rectangular slab is placed above the pipe and we will calculate the heat transfer as follows.

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  where,    A = area

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             \Delta T = change in temperature.

Therefore, putting the given values into the above formula as follows.

            Q = kA \frac{\Delta T}{L}

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Therefore, heat lost will be calculated as follows.

                       Q = kA \frac{\Delta T}{L}

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                           = 18.67 W

Thus, we can conclude that 18.67 W heat lost if the pipe was buried at a depth of 180 cm.

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A scuba tank contains a mixture of oxygen (O2) and nitrogen (N2) gas. The oxygen has a partial pressure of PO2=5.62MPa. The tota
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Answer:

21.16 MPa

Explanation:

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This implies that;

Total pressure of the system = partial pressure of nitrogen + partial pressure of oxygen

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Answer:

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