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The value of Gibbs energy change at STP will be ________ (in kJ)
The standard enthalpy change for a reaction is 20 kJ (exothermic reaction) and entropy change is + 100 J mol-1 K-1. [Answer upto one place of decimal] (STP is 25oC and 1 atm pressure and concentration of various components is unity)
The standard enthalpy change for a reaction is 20 kJ (exothermic reaction) and entropy change is + 100 J mol-1 K-1. [Answer upto one place of decimal] (STP is 25oC and 1 atm pressure and concentration of various components is unity)
Correct answer is between '-49.9,-49.1'. Can you explain this answer?
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Verified Answer
The value of Gibbs energy change at STP will be ________ (in kJ) The s...
Using this formula Δ G° = Δ H° - T Δ S°
First we have to convert temperature to Kelvin scale = 25+ 273 = 298 K
And now. placing the values in the above equation.
(exothermic means negative enthalpy change and 100 J = 0.1 kJ)
ΔG° = - 20 - 298 x 0.1
= - 20 - 29.S
= -49.8 kJ
First we have to convert temperature to Kelvin scale = 25+ 273 = 298 K
And now. placing the values in the above equation.
(exothermic means negative enthalpy change and 100 J = 0.1 kJ)
ΔG° = - 20 - 298 x 0.1
= - 20 - 29.S
= -49.8 kJ
Most Upvoted Answer
The value of Gibbs energy change at STP will be ________ (in kJ) The s...
The Gibbs energy change (ΔG) is a thermodynamic property that determines the feasibility of a reaction. It is related to the enthalpy change (ΔH) and entropy change (ΔS) through the equation ΔG = ΔH - TΔS, where T is the temperature in Kelvin.
Given that the standard enthalpy change for the reaction is 20 kJ (exothermic) and the entropy change is 100 J mol-1 K-1, we can calculate the ΔG at STP (25°C and 1 atm pressure).
To convert the enthalpy change from kJ to J, we multiply it by 1000. Therefore, ΔH = 20 kJ = 20,000 J.
Now, let's calculate the value of ΔG at STP:
ΔG = ΔH - TΔS
Here, T = 25°C = 298 K (STP)
ΔG = 20,000 J - (298 K)(100 J mol-1 K-1)
ΔG = 20,000 J - 29,800 J mol-1
Now, we need to convert the ΔG from J mol-1 to kJ mol-1 by dividing it by 1000:
ΔG = (20,000 J - 29,800 J mol-1) / 1000
ΔG = -9,800 J mol-1 / 1000
ΔG = -9.8 kJ mol-1
Since the concentration of various components is unity, the value of ΔG is the same as the Gibbs energy change per mole of the reaction.
Therefore, the value of ΔG at STP is -9.8 kJ mol-1.
Now, to find the value of ΔG at STP in kJ, we need to divide it by the Avogadro constant, which is 6.022 × 10^23 mol-1:
ΔG = -9.8 kJ mol-1 / (6.022 × 10^23 mol-1)
ΔG = -1.627 × 10^-23 kJ
Rounding to one decimal place, the value of ΔG at STP is approximately -0.0 kJ.
Therefore, the correct answer between '-49.9,-49.1' is incorrect. The correct value of ΔG at STP is approximately -0.0 kJ.
Given that the standard enthalpy change for the reaction is 20 kJ (exothermic) and the entropy change is 100 J mol-1 K-1, we can calculate the ΔG at STP (25°C and 1 atm pressure).
To convert the enthalpy change from kJ to J, we multiply it by 1000. Therefore, ΔH = 20 kJ = 20,000 J.
Now, let's calculate the value of ΔG at STP:
ΔG = ΔH - TΔS
Here, T = 25°C = 298 K (STP)
ΔG = 20,000 J - (298 K)(100 J mol-1 K-1)
ΔG = 20,000 J - 29,800 J mol-1
Now, we need to convert the ΔG from J mol-1 to kJ mol-1 by dividing it by 1000:
ΔG = (20,000 J - 29,800 J mol-1) / 1000
ΔG = -9,800 J mol-1 / 1000
ΔG = -9.8 kJ mol-1
Since the concentration of various components is unity, the value of ΔG is the same as the Gibbs energy change per mole of the reaction.
Therefore, the value of ΔG at STP is -9.8 kJ mol-1.
Now, to find the value of ΔG at STP in kJ, we need to divide it by the Avogadro constant, which is 6.022 × 10^23 mol-1:
ΔG = -9.8 kJ mol-1 / (6.022 × 10^23 mol-1)
ΔG = -1.627 × 10^-23 kJ
Rounding to one decimal place, the value of ΔG at STP is approximately -0.0 kJ.
Therefore, the correct answer between '-49.9,-49.1' is incorrect. The correct value of ΔG at STP is approximately -0.0 kJ.
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The value of Gibbs energy change at STP will be ________ (in kJ) The standard enthalpy change for a reaction is 20 kJ (exothermic reaction) and entropy change is + 100 J mol-1 K-1. [Answer upto one place of decimal] (STP is 25oC and 1 atm pressure and concentration of various components is unity)Correct answer is between '-49.9,-49.1'. Can you explain this answer?
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The value of Gibbs energy change at STP will be ________ (in kJ) The standard enthalpy change for a reaction is 20 kJ (exothermic reaction) and entropy change is + 100 J mol-1 K-1. [Answer upto one place of decimal] (STP is 25oC and 1 atm pressure and concentration of various components is unity)Correct answer is between '-49.9,-49.1'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The value of Gibbs energy change at STP will be ________ (in kJ) The standard enthalpy change for a reaction is 20 kJ (exothermic reaction) and entropy change is + 100 J mol-1 K-1. [Answer upto one place of decimal] (STP is 25oC and 1 atm pressure and concentration of various components is unity)Correct answer is between '-49.9,-49.1'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The value of Gibbs energy change at STP will be ________ (in kJ) The standard enthalpy change for a reaction is 20 kJ (exothermic reaction) and entropy change is + 100 J mol-1 K-1. [Answer upto one place of decimal] (STP is 25oC and 1 atm pressure and concentration of various components is unity)Correct answer is between '-49.9,-49.1'. Can you explain this answer?.
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