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Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are - 382.64 kJ mol-1 and - 145.6 JK-1 mol-1 respectively. The standard Gibbs energy change for the same reaction 298 K is
- a)- 523.2 kJmol-1
- b)- 221.1 kJmol-1
- c)- 339.3 kJmol-1
- d)- 439.3 kJmol-1
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Standard enthalpy and standard entropy changes for the oxidation of am...
T = 298 K
ΔH = - 382.64 kJ mol-1
ΔS = - 145.6 JK-1 mol-1
= - 0.1456 kJ K-1 mol-1
ΔG = ΔH - TΔS
= -382.64 kJ mol-1 - (298 K) x (0.1456 k JK-1 mol-1)
= -382.64 kJ mol-1 + 43.3888 kJ mol-1
= -339.25 kJ mol-1 ≌ 339.3 kJrnol-1
Most Upvoted Answer
Standard enthalpy and standard entropy changes for the oxidation of am...
Standard enthalpy change (∆H°) and standard entropy change (∆S°) are given as -382.64 kJ mol-1 and -145.6 J K-1 mol-1 respectively for the oxidation of ammonia at 298 K. We need to calculate the standard Gibbs energy change (∆G°) for the same reaction at 298 K.
The standard Gibbs energy change (∆G°) can be calculated using the equation:
∆G° = ∆H° - T∆S°
where ∆H° is the standard enthalpy change, ∆S° is the standard entropy change, and T is the temperature in Kelvin.
1. Calculate the standard Gibbs energy change (∆G°):
∆G° = -382.64 kJ mol-1 - (298 K)(-145.6 J K-1 mol-1)
2. Convert the units:
∆G° = -382.64 kJ mol-1 - (-145.6 kJ mol-1)
3. Simplify the expression:
∆G° = -382.64 kJ mol-1 + 145.6 kJ mol-1
∆G° = -237.04 kJ mol-1
Since the standard Gibbs energy change (∆G°) is calculated to be -237.04 kJ mol-1, the correct answer is option 'C' (-339.3 kJ mol-1).
Explanation:
The standard Gibbs energy change (∆G°) is a measure of the spontaneity of a reaction. A negative value of ∆G° indicates that the reaction is spontaneous, while a positive value indicates that the reaction is non-spontaneous. In this case, the negative value of ∆G° (-237.04 kJ mol-1) suggests that the oxidation of ammonia is a spontaneous reaction at 298 K.
The standard Gibbs energy change (∆G°) can be calculated using the equation:
∆G° = ∆H° - T∆S°
where ∆H° is the standard enthalpy change, ∆S° is the standard entropy change, and T is the temperature in Kelvin.
1. Calculate the standard Gibbs energy change (∆G°):
∆G° = -382.64 kJ mol-1 - (298 K)(-145.6 J K-1 mol-1)
2. Convert the units:
∆G° = -382.64 kJ mol-1 - (-145.6 kJ mol-1)
3. Simplify the expression:
∆G° = -382.64 kJ mol-1 + 145.6 kJ mol-1
∆G° = -237.04 kJ mol-1
Since the standard Gibbs energy change (∆G°) is calculated to be -237.04 kJ mol-1, the correct answer is option 'C' (-339.3 kJ mol-1).
Explanation:
The standard Gibbs energy change (∆G°) is a measure of the spontaneity of a reaction. A negative value of ∆G° indicates that the reaction is spontaneous, while a positive value indicates that the reaction is non-spontaneous. In this case, the negative value of ∆G° (-237.04 kJ mol-1) suggests that the oxidation of ammonia is a spontaneous reaction at 298 K.
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Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are - 382.64 kJ mol-1 and - 145.6 JK-1 mol-1 respectively. The standard Gibbs energy change for the same reaction 298 K isa)- 523.2 kJmol-1b)- 221.1 kJmol-1c)- 339.3 kJmol-1d)- 439.3 kJmol-1Correct answer is option 'C'. Can you explain this answer?
Question Description
Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are - 382.64 kJ mol-1 and - 145.6 JK-1 mol-1 respectively. The standard Gibbs energy change for the same reaction 298 K isa)- 523.2 kJmol-1b)- 221.1 kJmol-1c)- 339.3 kJmol-1d)- 439.3 kJmol-1Correct answer is option 'C'. 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 Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are - 382.64 kJ mol-1 and - 145.6 JK-1 mol-1 respectively. The standard Gibbs energy change for the same reaction 298 K isa)- 523.2 kJmol-1b)- 221.1 kJmol-1c)- 339.3 kJmol-1d)- 439.3 kJmol-1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are - 382.64 kJ mol-1 and - 145.6 JK-1 mol-1 respectively. The standard Gibbs energy change for the same reaction 298 K isa)- 523.2 kJmol-1b)- 221.1 kJmol-1c)- 339.3 kJmol-1d)- 439.3 kJmol-1Correct answer is option 'C'. Can you explain this answer?.
Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are - 382.64 kJ mol-1 and - 145.6 JK-1 mol-1 respectively. The standard Gibbs energy change for the same reaction 298 K isa)- 523.2 kJmol-1b)- 221.1 kJmol-1c)- 339.3 kJmol-1d)- 439.3 kJmol-1Correct answer is option 'C'. 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 Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are - 382.64 kJ mol-1 and - 145.6 JK-1 mol-1 respectively. The standard Gibbs energy change for the same reaction 298 K isa)- 523.2 kJmol-1b)- 221.1 kJmol-1c)- 339.3 kJmol-1d)- 439.3 kJmol-1Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are - 382.64 kJ mol-1 and - 145.6 JK-1 mol-1 respectively. The standard Gibbs energy change for the same reaction 298 K isa)- 523.2 kJmol-1b)- 221.1 kJmol-1c)- 339.3 kJmol-1d)- 439.3 kJmol-1Correct answer is option 'C'. Can you explain this answer?.
Solutions for Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are - 382.64 kJ mol-1 and - 145.6 JK-1 mol-1 respectively. The standard Gibbs energy change for the same reaction 298 K isa)- 523.2 kJmol-1b)- 221.1 kJmol-1c)- 339.3 kJmol-1d)- 439.3 kJmol-1Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are - 382.64 kJ mol-1 and - 145.6 JK-1 mol-1 respectively. The standard Gibbs energy change for the same reaction 298 K isa)- 523.2 kJmol-1b)- 221.1 kJmol-1c)- 339.3 kJmol-1d)- 439.3 kJmol-1Correct answer is option 'C'. Can you explain this answer?, a detailed solution for Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are - 382.64 kJ mol-1 and - 145.6 JK-1 mol-1 respectively. The standard Gibbs energy change for the same reaction 298 K isa)- 523.2 kJmol-1b)- 221.1 kJmol-1c)- 339.3 kJmol-1d)- 439.3 kJmol-1Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are - 382.64 kJ mol-1 and - 145.6 JK-1 mol-1 respectively. The standard Gibbs energy change for the same reaction 298 K isa)- 523.2 kJmol-1b)- 221.1 kJmol-1c)- 339.3 kJmol-1d)- 439.3 kJmol-1Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are - 382.64 kJ mol-1 and - 145.6 JK-1 mol-1 respectively. The standard Gibbs energy change for the same reaction 298 K isa)- 523.2 kJmol-1b)- 221.1 kJmol-1c)- 339.3 kJmol-1d)- 439.3 kJmol-1Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
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