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laila [671]
3 years ago
13

Under what conditions can a non-spontaneous reaction be made spontaneous?

Chemistry
2 answers:
Kitty [74]3 years ago
7 0
It can be c I think tho
Gre4nikov [31]3 years ago
6 0

Answer:

A. When an endothermic reaction has a positive change in entropy and the temperature is greatly increased

Explanation:

An spontaneous reaction is one that occurs without having been provoked by an external influence. The non-epontaneous reaction, on the other hand, is one that occurs only when they are induced in the opposite direction.

Most of the non-spontaneous reactions are endothermic, that is, they occur with the absorption of energy. An example of a non-spontaneous reaction occurs when a metal needs to be heated to increase its temperature. The heating of a metal is very difficult to occur naturally, without an external factor that provokes the reaction.

However, a non-spontaneous reaction can become spontaneous depending on the entropy of the system. Entropy in an isolated system tends to increase when an spontaneous reaction occurs. Therefore, in order for a non-spontaneous reaction to become spontaneous it is necessary that an endothermic reaction has a positive change in entropy and the temperature increases a lot.

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monitta

Answer:

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How many molecules of chlorine are needed to react with 56.Og of iron to form Iron (III) chloride (FeCl3)?

5 0
2 years ago
A 265-mL flask contains pure helium at a pressure of 751 torrs. A second flask with a volume of 465 mL contains pure argon at a
Nadya [2.5K]

Answer:

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar =  472.8 torr

Explanation:

Step 1: Data given

Volume of the flask helium = 265 mL

Pressure in the helium flask = 751 torr = 751/760 atm

Volume of the flask argon = 465 mL

Pressure in the argon flask = 727 torr = 727/760 atm

The total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual component in a gas mixture.

Step 2: Calculate total volume

Total volume = 265 mL + 465 mL = 730 mL =  0.730 L

Step 3: Boyle's Law:

P1V1=P2V2

⇒ with P1 = total pressure gas exerts in its own flask

 ⇒ with V1 = volume of flask with stopcock valve closed

 ⇒ with P2 = partial pressure of gas exerts on total volume of both flasks when stopcock valve is opened  

 ⇒ with V2 = total volume of both flasks with stopcock valve opened

Helium using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of helium = 751 /760 = 0.98816 atm

 ⇒ with V1 = volume of helium = 0.265 L

 ⇒ with P2 = The new partial pressure of helium

 ⇒ with V2 = total volume = 0.730 L

(0.98816 atm)(0.265L)=P2(0.730L)

P2=0.359 atm

Argon using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of argon = 727/760 = 0.95658 atm

 ⇒ with V1 = volume of argon = 0.465 L

 ⇒ with P2 = The new partial pressure of argon

 ⇒ with V2 = total volume = 0.730 L

(0.95658 atm)(0.465L)=P2(0.730L)

P2=0.609 atm

Step 4: Convert pressure in atm to torr

Pressure helium = 0.359 atm = 272.8 torr

Pressure argon = 0.609 atm = 472.8 torr

Step 5: Calculate Total pressure

Ptotal = P(He)+P(Ar)

⇒ Pt  = total pressure of the gas mixture

⇒ P(He) = partial pressure of Helium

 ⇒ P(Ar)  = partial pressure of Argon

Pt = 272.8 torr + 472.8 torr

Pt = 745.6 torr

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar =  472.8 torr

5 0
3 years ago
Find the number of moles of water that can be formed if you have 138 mol of hydrogen gas and 64 mol of oxygen gas.
charle [14.2K]
2H₂₍g₎ + O₂ ₍g₎→ 2H₂O

138 mol H₂ × (2 mol H₂O ÷ 2 mol H₂)= 138 mol H₂O
64 mol O₂ × (2 mol H₂O ÷ 1 mol O₂)= 128 mol H₂O

128 mol H₂O
6 0
3 years ago
Read 2 more answers
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Zolol [24]

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Explanation:

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3 years ago
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Calculate the theoretical yield of ammonia produced by the reaction of 100 g of H2 gas and 200g of N2 gas? 3H2(g) + N2(g)-----&g
Alex787 [66]
Moles of Hydrogen present: 100 / 2 = 50 moles

Moles of Nitrogen present: 200 / 28 = 7.14 moles

Hydrogen required by given amount of nitrogen = 7.14 x 3 = 21.42 moles

Hydrogen is excess so we will calculate the Ammonia produced using Nitrogen.

Molar ratio of Nitrogen : Ammonia = 1 : 2

Moles of ammonia = 7.14 x 2 = 14.28 moles
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3 years ago
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