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SCORPION-xisa [38]
3 years ago
5

What happens when carbon dioxide at 74°C is placed in thermal contact with water at 14°C? Energy leaves the carbon dioxide and e

nters the water. The water gets hotter than the carbon dioxide. The temperatures of the two substances equalize. Both the water and the carbon dioxide decrease in temperature. The water stays colder than the carbon dioxide.
Chemistry
1 answer:
qaws [65]3 years ago
5 0
<h3><u>Answer;</u></h3>

The temperatures of the two substances equalize.

<h3><u>Explanation;</u></h3>
  • When two objects at different temperatures are placed in contact, heat flows from the warmer object to the cooler one until the temperature of both objects is the same.
  • <em><u>The amount of heat that flows from a warmer object is the same as the amount of heat that flows into a cooler object. Because the direction of heat flow is opposite for the two objects, the sign of the heat flow values must be opposite. </u></em>
  • <em><u>In line with the law of conservation of energy, the amount of heat lost by a warmer object equals the amount of heat gained by a cooler object. Therefore the amount of heat lost by carbon dioxide is equal to the amount of heat gained by water.</u></em>
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Part A
Elden [556K]

Answer:

ΔG° = -5.4 kJ/mol

ΔG = 873.2 J/mol = 0.873 kJ /mol

Explanation:

Step 1: Data given

ΔG (NO2) = 51.84 kJ/mol

ΔG (N2O4)  = 98.28 kJ/mol

Step 2:

ΔG = ΔG° + RT ln Q

⇒with Q = the reaction quatient

⇒with T = the temperature = 298 K

⇒with R = 8.314 J / mol*K

⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2 )

⇒ ΔG° = 98.28 kJ/mol - 2* 51.84  kJ/mol

⇒ ΔG° = -5.4 kJ/mol

Part B

ΔG =  ΔG° =RT ln Q

⇒with G° = -5.4 kj/mol = -5400 j/mol

⇒ with R = 8.314 J/K*mol

⇒with T = 298 K

⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577

ΔG = -5400 + 8.314 * 298 * ln(12.577)

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This is a solubility equilibrium, where you have a precipitate formed, lead(II) chloride. This salt can be dissociated as:

           PbCl₂(s)  ⇄  Pb²⁺ (aq)  +  2Cl⁻ (aq)     Kps

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Eq          x - s              s                  2s

As this is an equilibrium, the Kps works as the constant (Solubility product):

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