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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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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?
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