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shusha [124]
2 years ago
7

Questions 18-20 refer to the reaction 2NH3 (g) ↔ N2 (g) + 3H2 (g), and ΔH = 92.2 kJ.

Chemistry
1 answer:
Fittoniya [83]2 years ago
4 0
18. <span>Answer is </span>A

<span>  
<span>Since the enthalpy of reaction is positive, the forward reaction is<span> an endothermic reaction which means the energy is gained from the surrounding to happen the reaction. If the temperature decreases, according to the </span></span>Le Chatelier's principle, the system tries to become equilibrium by increasing temperature. Since forward reaction is endothermic (because of the bond breaking), the backward reaction is exothermic (because of the bond making) which releases the energy to the surroundings. This makes the increase of temperature. So if the backward reaction is promoted because of the decrease of temperature, then the concentration of H</span><span>₂ will decrease.</span>

<span>
</span>

19. Answer is A.

The reactant side has 2 moles/molecules of reactants and the product side has 4 moles/molecules of products which come from 1 N₂(g) and 3 H₂<span>(g). If the pressure is reduced in the system, according to the Le Chatelier's principle, the system tries to increase the pressure. </span><span>Hence, forward reaction is promoted because of the higher number of molecules in product side. If the forward reaction is promoted, the concentration of NH</span>₃(g) will decreased.


<span>20. </span>Answer is C.

If the concentration of reactant is increased in the system, according to the Le Chatelier's principle, the system tries to reduce the concentration of that reactant. So if NH₃(g) concentration is increased, then to be equilibrium, the forward reaction will be promoted. Then the concentration of N₂<span>(g) will increase.</span>

<span> </span>

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How many joules are required to melt 100 grams of water? The heat of
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a)BH3.THF is used to convert 1-pentane to pentanoic acid and b)NaCN is used to convert Bromobutane to pentanoic acid.

a) The conversion of 1-pentane to pentanoic acid using BH3, also known as hydroboration-oxidation, is a two-step reaction involving the reaction of 1-pentane with borane (BH3), followed by oxidation of the resulting 1-pentylborane with hydrogen peroxide or other oxidizing agents.

In the first step, 1-pentane reacts with borane (BH3) to form 1-pentylborane, through a process known as hydroboration. This reaction is catalyzed by a Lewis acid, such as aluminum chloride, and proceeds via a hydride transfer from the borane to the 1-pentane.

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The overall chemical reaction for the conversion of 1-pentane to pentanoic acid using borane (BH₃) and hydrogen peroxide (H₂O₂) is as follows:

1-pentane + BH₃ + H₂O₂ → pentanoic acid + H₂O + BH₂

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2. Loss of a proton from the tetrahedral intermediate to form a carbanion.

3. Protonation of the carbanion by water (or another proton source) to form pentanoic acid.

The overall reaction can be represented as follows:

1-Bromo butane + NaCN → Pentanoic Acid + NaBr

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