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docker41 [41]
4 years ago
7

For the vaporization of a liquid at a given pressure: A.ΔG is positive at all temperatures. B.ΔG is negative at all temperatures

. C.ΔG is positive at low temperatures, but negative at high temperatures (and zero at some temperature). D.ΔG is negative at low temperatures, but positive at high temperatures (and zero at some temperature). E.None of these.
Chemistry
1 answer:
MArishka [77]4 years ago
5 0

Answer:

C. ΔG is positive at low temperatures, but negative at high temperatures (and zero at some temperature).

Explanation:

Since we need to give energy in the form of heat to vaporize a liquid, the enthalpy is positive. In a gas, molecules are more separated than in a liquid, therefore the entropy is positive as well.

Considering the Gibbs free energy equation:

ΔG= ΔH - TΔS

          +        +

When both the enthalpy and entropy are positive, the reaction proceeds spontaneously (ΔG is negative) at high temperatures. At low temperatures, the reaction is spontaneous in the reverse direction (ΔG is positive).

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true

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pressure has little effect on the solubility of liquids and solids because they are almost incompressible. True or False
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Pressure has little effect on the solubility of liquids and solids because they are almost incompressible True.

Liquids and solids show little change in solubility with changes in pressure. As expected, gases increase in solubility with increasing pressure. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the pressure of that gas above the surface of the solution.

External pressure has little effect on liquid and solid solubility. In contrast, the solubility of a gas increases as the partial pressure of the gas above the solution increases.

Solubility is a measure of the concentration of dissolved gas particles in a liquid and is a function of gas pressure. Increasing the gas pressure increases the number of collisions and increases the solubility, and decreasing the pressure decreases the solubility.

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1 year ago
Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cr3+(aq) + Pb(s)2Cr2+(aq) + Pb2+(aq)
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Answer:

3.47 ×10^-10

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The equation of the reaction is 2Cr3+(aq) + Pb(s)------->2Cr2+(aq) + Pb2+(aq)

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E°cell = E°cathode - E°anode

E°cathode = -0.41 V

E°anode = -0.13 V

E°cell = -0.41 -(-0.13) = -0.28 V

From

E°cell = 0.0592/n log K

n= 2, K= the unknown

-0.28 = 0.0592/2 log K

log K = -0.28/0.0296

log K = -9.4595

K = Antilog ( -9.4595)

K= 3.47 ×10^-10

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