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When enthalpy and entropy change for a chemical reaction are -2.5 x103 cals and 7.4 cals deg⁻1 respectively . Predict that reaction at 298 K is
- a)spontaneous
- b)reversible
- c)irreversible
- d)non-spontaneous
Correct answer is option 'A'. Can you explain this answer?
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When enthalpy and entropy change for a chemical reaction are -2.5 x103...
Most Upvoted Answer
When enthalpy and entropy change for a chemical reaction are -2.5 x103...
Given data:
Enthalpy change, ΔH = -2.5 x 10³ cal
Entropy change, ΔS = 7.4 cal/°C
Temperature, T = 298 K
To determine the spontaneity of the reaction, we can use the Gibbs free energy change (ΔG) of the reaction.
ΔG = ΔH - TΔS
Substituting the given values:
ΔG = -2.5 x 10³ - 298 x 7.4 = -2.5 x 10³ - 2205.2 = -2207.7 cal
If ΔG is negative, the reaction will be spontaneous.
Since ΔG is negative, the reaction is spontaneous at 298K.
Therefore, the correct answer is option 'A' - spontaneous.
Enthalpy change, ΔH = -2.5 x 10³ cal
Entropy change, ΔS = 7.4 cal/°C
Temperature, T = 298 K
To determine the spontaneity of the reaction, we can use the Gibbs free energy change (ΔG) of the reaction.
ΔG = ΔH - TΔS
Substituting the given values:
ΔG = -2.5 x 10³ - 298 x 7.4 = -2.5 x 10³ - 2205.2 = -2207.7 cal
If ΔG is negative, the reaction will be spontaneous.
Since ΔG is negative, the reaction is spontaneous at 298K.
Therefore, the correct answer is option 'A' - spontaneous.
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Community Answer
When enthalpy and entropy change for a chemical reaction are -2.5 x103...
Given data:
Enthalpy change (ΔH) = -2.5 x 103 cal
Entropy change (ΔS) = 7.4 cal/deg
To determine the spontaneity of the reaction, we can use the Gibbs free energy change (ΔG) which is given by the equation:
ΔG = ΔH - TΔS
where T is the temperature in Kelvin.
Now let's substitute the given values in the above equation:
ΔG = -2.5 x 103 - (298)(7.4)
ΔG = -2.5 x 103 - 2205.2
ΔG = -2705.2 cal
Since the value of ΔG is negative, the reaction is spontaneous at 298 K.
Therefore, the correct option is (a) spontaneous.
Enthalpy change (ΔH) = -2.5 x 103 cal
Entropy change (ΔS) = 7.4 cal/deg
To determine the spontaneity of the reaction, we can use the Gibbs free energy change (ΔG) which is given by the equation:
ΔG = ΔH - TΔS
where T is the temperature in Kelvin.
Now let's substitute the given values in the above equation:
ΔG = -2.5 x 103 - (298)(7.4)
ΔG = -2.5 x 103 - 2205.2
ΔG = -2705.2 cal
Since the value of ΔG is negative, the reaction is spontaneous at 298 K.
Therefore, the correct option is (a) spontaneous.
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When enthalpy and entropy change for a chemical reaction are -2.5 x103 cals and 7.4 cals deg1 respectively . Predict that reaction at 298 K isa)spontaneousb)reversiblec)irreversibled)non-spontaneousCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about When enthalpy and entropy change for a chemical reaction are -2.5 x103 cals and 7.4 cals deg1 respectively . Predict that reaction at 298 K isa)spontaneousb)reversiblec)irreversibled)non-spontaneousCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When enthalpy and entropy change for a chemical reaction are -2.5 x103 cals and 7.4 cals deg1 respectively . Predict that reaction at 298 K isa)spontaneousb)reversiblec)irreversibled)non-spontaneousCorrect answer is option 'A'. Can you explain this answer?.
When enthalpy and entropy change for a chemical reaction are -2.5 x103 cals and 7.4 cals deg1 respectively . Predict that reaction at 298 K isa)spontaneousb)reversiblec)irreversibled)non-spontaneousCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about When enthalpy and entropy change for a chemical reaction are -2.5 x103 cals and 7.4 cals deg1 respectively . Predict that reaction at 298 K isa)spontaneousb)reversiblec)irreversibled)non-spontaneousCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When enthalpy and entropy change for a chemical reaction are -2.5 x103 cals and 7.4 cals deg1 respectively . Predict that reaction at 298 K isa)spontaneousb)reversiblec)irreversibled)non-spontaneousCorrect answer is option 'A'. Can you explain this answer?.
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