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The contributions of both heat (enthalpy) and randomness (entropy) shall be considered to the overall spontaneity of a process. When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibb's energy change (ΔG).
ΔG = ΔH – TΔS
where, ΔH = Enthalpy change ; ΔS = Entropy change;
T = Temperature in kelvin.
If ΔG < 0, Process is spontaneous ; ΔG = 0, Process is at equilibrium, AG > 0, Process is non-spontaneous.
Q.
For the change
H20 (s, 273 K, 2 atm) → H20 (l, 273K, 2 atom),
choose the correct option.
- a)AG = 0
- b)AG < 0
- c)AG > 0 -
- d)None
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The contributions of both heat (enthalpy) and randomness (entropy) sha...
H2O (s) →H2O ( l ),
As P , more H2O ( l ) is formed.
At P = 1 atm, Δ G = 0.
At P = 1 atm, Δ G = 0.
Most Upvoted Answer
The contributions of both heat (enthalpy) and randomness (entropy) sha...
Explanation:
To determine the spontaneity of the process, we need to calculate the Gibbs energy change (ΔG) using the formula:
ΔG = ΔH - TΔS
Given that the reaction is H2O (s, 273 K, 2 atm) to H2O (l, 273 K, 2 atm), we can break down the calculation as follows:
Enthalpy Change (ΔH):
The enthalpy change (ΔH) represents the heat transferred during the process. In this case, the enthalpy change for the phase transition from solid to liquid water is the heat of fusion, which is positive. So, ΔH > 0.
Entropy Change (ΔS):
The entropy change (ΔS) represents the change in randomness or disorder of the system. In this case, the entropy change for the phase transition from solid to liquid water is positive. The transition from a solid to a liquid state generally results in an increase in disorder. So, ΔS > 0.
Temperature (T):
The temperature (T) is given as 273 K.
Calculation of ΔG:
Using the formula ΔG = ΔH - TΔS, we can substitute the values:
ΔG = ΔH - TΔS
ΔG = ΔH - (273 K)(ΔS)
Since ΔH and ΔS are both positive, the first term ΔH is positive and the second term (273 K)(ΔS) is also positive. Therefore, ΔG will be negative.
Conclusion:
Since ΔG is negative (ΔG < 0),="" the="" process="" is="" spontaneous.="" therefore,="" the="" correct="" answer="" is="" option="" 'b'="" (δg="" />< 0).="" />
To determine the spontaneity of the process, we need to calculate the Gibbs energy change (ΔG) using the formula:
ΔG = ΔH - TΔS
Given that the reaction is H2O (s, 273 K, 2 atm) to H2O (l, 273 K, 2 atm), we can break down the calculation as follows:
Enthalpy Change (ΔH):
The enthalpy change (ΔH) represents the heat transferred during the process. In this case, the enthalpy change for the phase transition from solid to liquid water is the heat of fusion, which is positive. So, ΔH > 0.
Entropy Change (ΔS):
The entropy change (ΔS) represents the change in randomness or disorder of the system. In this case, the entropy change for the phase transition from solid to liquid water is positive. The transition from a solid to a liquid state generally results in an increase in disorder. So, ΔS > 0.
Temperature (T):
The temperature (T) is given as 273 K.
Calculation of ΔG:
Using the formula ΔG = ΔH - TΔS, we can substitute the values:
ΔG = ΔH - TΔS
ΔG = ΔH - (273 K)(ΔS)
Since ΔH and ΔS are both positive, the first term ΔH is positive and the second term (273 K)(ΔS) is also positive. Therefore, ΔG will be negative.
Conclusion:
Since ΔG is negative (ΔG < 0),="" the="" process="" is="" spontaneous.="" therefore,="" the="" correct="" answer="" is="" option="" 'b'="" (δg="" />< 0).="" />
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The contributions of both heat (enthalpy) and randomness (entropy) shall be considered to the overall spontaneity of a process. When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibbs energy change (G).G = H TSwhere,H = Enthalpy change ; S = Entropy change;T = Temperature in kelvin.If
G 0, Process is spontaneous ;
G = 0, Process is at equilibrium, AG 0, Process is non-spontaneous.Q.For the changeH20 (s, 273 K, 2 atm) H20 (l, 273K, 2 atom),choose the correct option.a)AG = 0b)AG 0c)AG 0 -d)NoneCorrect answer is option 'B'. Can you explain this answer?
Question Description
The contributions of both heat (enthalpy) and randomness (entropy) shall be considered to the overall spontaneity of a process. When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibbs energy change (G).G = H TSwhere,H = Enthalpy change ; S = Entropy change;T = Temperature in kelvin.If G 0, Process is spontaneous ; G = 0, Process is at equilibrium, AG 0, Process is non-spontaneous.Q.For the changeH20 (s, 273 K, 2 atm) H20 (l, 273K, 2 atom),choose the correct option.a)AG = 0b)AG 0c)AG 0 -d)NoneCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The contributions of both heat (enthalpy) and randomness (entropy) shall be considered to the overall spontaneity of a process. When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibbs energy change (G).G = H TSwhere,H = Enthalpy change ; S = Entropy change;T = Temperature in kelvin.If G 0, Process is spontaneous ; G = 0, Process is at equilibrium, AG 0, Process is non-spontaneous.Q.For the changeH20 (s, 273 K, 2 atm) H20 (l, 273K, 2 atom),choose the correct option.a)AG = 0b)AG 0c)AG 0 -d)NoneCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The contributions of both heat (enthalpy) and randomness (entropy) shall be considered to the overall spontaneity of a process. When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibbs energy change (G).G = H TSwhere,H = Enthalpy change ; S = Entropy change;T = Temperature in kelvin.If G 0, Process is spontaneous ; G = 0, Process is at equilibrium, AG 0, Process is non-spontaneous.Q.For the changeH20 (s, 273 K, 2 atm) H20 (l, 273K, 2 atom),choose the correct option.a)AG = 0b)AG 0c)AG 0 -d)NoneCorrect answer is option 'B'. Can you explain this answer?.
The contributions of both heat (enthalpy) and randomness (entropy) shall be considered to the overall spontaneity of a process. When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibbs energy change (G).G = H TSwhere,H = Enthalpy change ; S = Entropy change;T = Temperature in kelvin.If G 0, Process is spontaneous ; G = 0, Process is at equilibrium, AG 0, Process is non-spontaneous.Q.For the changeH20 (s, 273 K, 2 atm) H20 (l, 273K, 2 atom),choose the correct option.a)AG = 0b)AG 0c)AG 0 -d)NoneCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The contributions of both heat (enthalpy) and randomness (entropy) shall be considered to the overall spontaneity of a process. When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibbs energy change (G).G = H TSwhere,H = Enthalpy change ; S = Entropy change;T = Temperature in kelvin.If G 0, Process is spontaneous ; G = 0, Process is at equilibrium, AG 0, Process is non-spontaneous.Q.For the changeH20 (s, 273 K, 2 atm) H20 (l, 273K, 2 atom),choose the correct option.a)AG = 0b)AG 0c)AG 0 -d)NoneCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The contributions of both heat (enthalpy) and randomness (entropy) shall be considered to the overall spontaneity of a process. When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibbs energy change (G).G = H TSwhere,H = Enthalpy change ; S = Entropy change;T = Temperature in kelvin.If G 0, Process is spontaneous ; G = 0, Process is at equilibrium, AG 0, Process is non-spontaneous.Q.For the changeH20 (s, 273 K, 2 atm) H20 (l, 273K, 2 atom),choose the correct option.a)AG = 0b)AG 0c)AG 0 -d)NoneCorrect answer is option 'B'. Can you explain this answer?.
Solutions for The contributions of both heat (enthalpy) and randomness (entropy) shall be considered to the overall spontaneity of a process. When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibbs energy change (G).G = H TSwhere,H = Enthalpy change ; S = Entropy change;T = Temperature in kelvin.If
G 0, Process is spontaneous ;
G = 0, Process is at equilibrium, AG 0, Process is non-spontaneous.Q.For the changeH20 (s, 273 K, 2 atm) H20 (l, 273K, 2 atom),choose the correct option.a)AG = 0b)AG 0c)AG 0 -d)NoneCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Here you can find the meaning of The contributions of both heat (enthalpy) and randomness (entropy) shall be considered to the overall spontaneity of a process. When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibbs energy change (G).G = H TSwhere,H = Enthalpy change ; S = Entropy change;T = Temperature in kelvin.If
G 0, Process is spontaneous ;
G = 0, Process is at equilibrium, AG 0, Process is non-spontaneous.Q.For the changeH20 (s, 273 K, 2 atm) H20 (l, 273K, 2 atom),choose the correct option.a)AG = 0b)AG 0c)AG 0 -d)NoneCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
The contributions of both heat (enthalpy) and randomness (entropy) shall be considered to the overall spontaneity of a process. When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibbs energy change (G).G = H TSwhere,H = Enthalpy change ; S = Entropy change;T = Temperature in kelvin.If
G 0, Process is spontaneous ;
G = 0, Process is at equilibrium, AG 0, Process is non-spontaneous.Q.For the changeH20 (s, 273 K, 2 atm) H20 (l, 273K, 2 atom),choose the correct option.a)AG = 0b)AG 0c)AG 0 -d)NoneCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for The contributions of both heat (enthalpy) and randomness (entropy) shall be considered to the overall spontaneity of a process. When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibbs energy change (G).G = H TSwhere,H = Enthalpy change ; S = Entropy change;T = Temperature in kelvin.If
G 0, Process is spontaneous ;
G = 0, Process is at equilibrium, AG 0, Process is non-spontaneous.Q.For the changeH20 (s, 273 K, 2 atm) H20 (l, 273K, 2 atom),choose the correct option.a)AG = 0b)AG 0c)AG 0 -d)NoneCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of The contributions of both heat (enthalpy) and randomness (entropy) shall be considered to the overall spontaneity of a process. When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibbs energy change (G).G = H TSwhere,H = Enthalpy change ; S = Entropy change;T = Temperature in kelvin.If
G 0, Process is spontaneous ;
G = 0, Process is at equilibrium, AG 0, Process is non-spontaneous.Q.For the changeH20 (s, 273 K, 2 atm) H20 (l, 273K, 2 atom),choose the correct option.a)AG = 0b)AG 0c)AG 0 -d)NoneCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice The contributions of both heat (enthalpy) and randomness (entropy) shall be considered to the overall spontaneity of a process. When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibbs energy change (G).G = H TSwhere,H = Enthalpy change ; S = Entropy change;T = Temperature in kelvin.If
G 0, Process is spontaneous ;
G = 0, Process is at equilibrium, AG 0, Process is non-spontaneous.Q.For the changeH20 (s, 273 K, 2 atm) H20 (l, 273K, 2 atom),choose the correct option.a)AG = 0b)AG 0c)AG 0 -d)NoneCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.
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