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Ugo [173]
3 years ago
5

Urea spontaneous dissolves in water (to a high concentration). When urea dissolves in water, the solution cools before reaching

thermal equilibrium with its surroundings. What can you say about the direction of change in the solution's entropy
Chemistry
1 answer:
frez [133]3 years ago
4 0

Answer:

  • <u>If the amount of heat released by the solution, when it cools, is greater than the decrease in the free energy, the entropy will decrease, and if the amount of heat released by the solution is less than the decrease in the free energy, the entropy decreases.</u>

Explanation:

In an spontaneus reaction the free energy decreases: ΔG < 0.

Since, the solution cools its enthalpy decreases: ΔH < 0.

By definition ΔG = ΔH - TΔS

Then, you can solve for TΔS:

  • TΔS = ΔH - ΔG

Since ΔG is negative, - ΔG is positive.

Then, TΔS will be a negative value (ΔH) plus a positive value (-ΔG).

If the magnitue of the negative value is greater than the positive value, TΔS will be negative, and the entropy decreases.

If, on the other hand, the magnitude of the negative value is less than the positive value, TΔS will be positive, and the entropy increases.

That means, that if the amount of heat released by the solution, when it cools, is greater than the decrease in the free energy, the entropy will decrease, and if the amount of heat released by the solution is less than the decrease in the free energy, the entropy decreases.

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Janet hikes a trail at a local forest each day. The trail is 3.6 miles long, and she hiked five days in the past week. How many
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4 years ago
During studies of the following reaction (i), a chemical engineer measured a less-than-expected yield of N2 and discovered that
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Answer:

Maximum expected yield = 87.2%

Explanation:

Equations of reactions:

Main reaction: N₂O₄(l) + 2N₂H₄(l) ---> 3N₂(g) + 4H₂O(g)

Side reaction:  2N₂O₄(l) + N₂H₄(l) ----> 6NO(g) + 2H₂O(g)

Molar mass of N₂O₄ = 92 g/mol; molar mass of N₂H₄ = 32 g/mol; molar mass of N₂ = 28 g/mol; molar mass of of NO = 30 g/mol; molar mass of water = 18 g/mol

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂.

101.1 g of N₂O₄ will react with 2 * 32 * 101.1 / 92 g of N₂H₄ = 70.33 g of N₂H₄

<em>N₂O₄ is the limiting reactant</em>

101.1 g of N₂O₄ will react to produce 3 * 14 * 101.1 / 92 g of N₂ =  46.15 g of  N₂

In the side reaction, (6 * 30 g) of NO  is produced  from (2 * 92 g) of N₂O₄ and 32 g of N₂H₄

12.7 g of N₂O₄ will be produced from ( 2 * 92 * 12.7/180 g) of N₂O₄ and (32 * 12.7/180) g of N₂H₄ to produce

mass of N₂O₄ used = 12.98 g

mass of N₂H₄ used = 2.26 g

mass of N₂O₄ left for main reaction = 101.1 - 12.98 = 88.12 g

mass of N₂H₄ left for main reaction = 101.1 - 2.26 = 98.84 g

In the main reaction, 92 g of N₂O₄ reacts with 2 * 32 g of N₂H₄ to produce 3 * 14 g of N₂

88.12 g of N₂O₄ will react with 2 * 32 * 88.12 / 92 g of N₂H₄ = 61.30 g of N₂H₄

N₂O₄ is the limiting reactant.

88.12 g of N₂O₄ will to react produce 3 * 14 * 88.12 / 92 g of N₂ =  40.23 g of  N₂

Percentage yield = (theoretical yield/actual yield) * 100%

Percentage yield = (40.23/46.15) * 100% = 87.2%

Therefore, maximum expected yield = 87.2%

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