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One mole of ice is converted into water at 273 K. The entropies of H2O(s) and H2O(l) are 38.20 and 60.01 J mol-1 K-1 respectively. The enthalpy change for the conversion is:
- a)59.54 J mol-1
- b)5954 J mol-1
- c)595.4 J mol-1
- d)320. 6 J mol-1
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
One mole of ice is converted into water at 273 K. The entropies of H2O...
or ΔH = 273 * (60.1 - 38.20) = 5954.13 J mol-1
Most Upvoted Answer
One mole of ice is converted into water at 273 K. The entropies of H2O...
Given:
- Number of moles of ice = 1 mole
- Temperature = 273 K
- Entropy of H2O(s) = 38.20 J mol-1 K-1
- Entropy of H2O(l) = 60.01 J mol-1 K-1
To Find:
Enthalpy change for the conversion
Solution:
Step 1: Calculate the change in entropy (ΔS)
The change in entropy can be calculated using the formula:
ΔS = S(l) - S(s)
where S(l) is the entropy of the liquid phase and S(s) is the entropy of the solid phase.
Given:
S(l) = 60.01 J mol-1 K-1
S(s) = 38.20 J mol-1 K-1
So, the change in entropy (ΔS) = 60.01 - 38.20 = 21.81 J mol-1 K-1
Step 2: Calculate the enthalpy change (ΔH)
The enthalpy change can be calculated using the formula:
ΔH = TΔS
where T is the temperature in Kelvin and ΔS is the change in entropy.
Given:
T = 273 K
ΔS = 21.81 J mol-1 K-1
So, the enthalpy change (ΔH) = 273 * 21.81 = 5954 J mol-1
Answer:
The enthalpy change for the conversion of one mole of ice into water at 273 K is 5954 J mol-1. Therefore, the correct answer is option B.
- Number of moles of ice = 1 mole
- Temperature = 273 K
- Entropy of H2O(s) = 38.20 J mol-1 K-1
- Entropy of H2O(l) = 60.01 J mol-1 K-1
To Find:
Enthalpy change for the conversion
Solution:
Step 1: Calculate the change in entropy (ΔS)
The change in entropy can be calculated using the formula:
ΔS = S(l) - S(s)
where S(l) is the entropy of the liquid phase and S(s) is the entropy of the solid phase.
Given:
S(l) = 60.01 J mol-1 K-1
S(s) = 38.20 J mol-1 K-1
So, the change in entropy (ΔS) = 60.01 - 38.20 = 21.81 J mol-1 K-1
Step 2: Calculate the enthalpy change (ΔH)
The enthalpy change can be calculated using the formula:
ΔH = TΔS
where T is the temperature in Kelvin and ΔS is the change in entropy.
Given:
T = 273 K
ΔS = 21.81 J mol-1 K-1
So, the enthalpy change (ΔH) = 273 * 21.81 = 5954 J mol-1
Answer:
The enthalpy change for the conversion of one mole of ice into water at 273 K is 5954 J mol-1. Therefore, the correct answer is option B.
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One mole of ice is converted into water at 273 K. The entropies of H2O(s) and H2O(l) are 38.20 and 60.01 J mol-1 K-1 respectively. The enthalpy change for the conversion is: a)59.54 J mol-1 b)5954 J mol-1 c)595.4 J mol-1 d)320. 6 J mol-1Correct answer is option 'B'. Can you explain this answer?
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One mole of ice is converted into water at 273 K. The entropies of H2O(s) and H2O(l) are 38.20 and 60.01 J mol-1 K-1 respectively. The enthalpy change for the conversion is: a)59.54 J mol-1 b)5954 J mol-1 c)595.4 J mol-1 d)320. 6 J mol-1Correct answer is option 'B'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about One mole of ice is converted into water at 273 K. The entropies of H2O(s) and H2O(l) are 38.20 and 60.01 J mol-1 K-1 respectively. The enthalpy change for the conversion is: a)59.54 J mol-1 b)5954 J mol-1 c)595.4 J mol-1 d)320. 6 J mol-1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of ice is converted into water at 273 K. The entropies of H2O(s) and H2O(l) are 38.20 and 60.01 J mol-1 K-1 respectively. The enthalpy change for the conversion is: a)59.54 J mol-1 b)5954 J mol-1 c)595.4 J mol-1 d)320. 6 J mol-1Correct answer is option 'B'. Can you explain this answer?.
One mole of ice is converted into water at 273 K. The entropies of H2O(s) and H2O(l) are 38.20 and 60.01 J mol-1 K-1 respectively. The enthalpy change for the conversion is: a)59.54 J mol-1 b)5954 J mol-1 c)595.4 J mol-1 d)320. 6 J mol-1Correct answer is option 'B'. Can you explain this answer? for Class 12 2025 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about One mole of ice is converted into water at 273 K. The entropies of H2O(s) and H2O(l) are 38.20 and 60.01 J mol-1 K-1 respectively. The enthalpy change for the conversion is: a)59.54 J mol-1 b)5954 J mol-1 c)595.4 J mol-1 d)320. 6 J mol-1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of ice is converted into water at 273 K. The entropies of H2O(s) and H2O(l) are 38.20 and 60.01 J mol-1 K-1 respectively. The enthalpy change for the conversion is: a)59.54 J mol-1 b)5954 J mol-1 c)595.4 J mol-1 d)320. 6 J mol-1Correct answer is option 'B'. Can you explain this answer?.
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One mole of ice is converted into water at 273 K. The entropies of H2O(s) and H2O(l) are 38.20 and 60.01 J mol-1 K-1 respectively. The enthalpy change for the conversion is: a)59.54 J mol-1 b)5954 J mol-1 c)595.4 J mol-1 d)320. 6 J mol-1Correct answer is option 'B'. Can you explain this answer?, a detailed solution for One mole of ice is converted into water at 273 K. The entropies of H2O(s) and H2O(l) are 38.20 and 60.01 J mol-1 K-1 respectively. The enthalpy change for the conversion is: a)59.54 J mol-1 b)5954 J mol-1 c)595.4 J mol-1 d)320. 6 J mol-1Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of One mole of ice is converted into water at 273 K. The entropies of H2O(s) and H2O(l) are 38.20 and 60.01 J mol-1 K-1 respectively. The enthalpy change for the conversion is: a)59.54 J mol-1 b)5954 J mol-1 c)595.4 J mol-1 d)320. 6 J mol-1Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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