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Given, ΔfH° of HCI (g) is - 22 . 10 kcal mol-1 and ΔSolutionH° (heat of solution) of HCI (g) is - 17.9 kcal mol-1. Thus, ΔfH ° of Cl- (aq) is
- a)-40.0 kcal mol-1
- b)+40.0 kcal mol-1
- c)+17.9 kcal mol-1
- d)+22.1 kcal mol-1
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Given, ΔfH° of HCI (g) is - 22 . 10 kcal mol-1 and ΔSol...
½ H2(g) + ½ Cl2(g) → HCl(g) ∆fH° = -22.10 kcal mol-1 -----(I)
HCl(g) + H2O(l) → H+(aq) + Cl-(aq) ∆solutionH° = -17.9 kcal mol-1 -----(II)
Let us add (I) and (II), we get
½ H2(g) + ½ Cl2(g) → H+(aq) + Cl-(aq) ∆rH° = -40.0 kcal mol-1
∆rH° = ∑∆Hproduct - ∑∆Hreactant = ∆fHCl- (since enthalpy of formation of H, H2 and Cl2 are 0)
∆fHCl- = -40.0
HCl(g) + H2O(l) → H+(aq) + Cl-(aq) ∆solutionH° = -17.9 kcal mol-1 -----(II)
Let us add (I) and (II), we get
½ H2(g) + ½ Cl2(g) → H+(aq) + Cl-(aq) ∆rH° = -40.0 kcal mol-1
∆rH° = ∑∆Hproduct - ∑∆Hreactant = ∆fHCl- (since enthalpy of formation of H, H2 and Cl2 are 0)
∆fHCl- = -40.0
Most Upvoted Answer
Given, ΔfH° of HCI (g) is - 22 . 10 kcal mol-1 and ΔSol...
Explanation:
Given Data:
- fH of HCl (g) = -22.10 kcal mol-1
- SolutionH of HCl (g) = -17.9 kcal mol-1
Calculating fH of Cl-(aq):
- The formation of Cl-(aq) from HCl (g) involves the dissociation of HCl into H+ and Cl- ions.
- The enthalpy change for this process is equal to the sum of the enthalpies of formation of the products minus the enthalpy of formation of the reactants.
- The enthalpy of formation of H+ is 0 kcal mol-1 as it is a free ion in solution.
- Therefore, fH of Cl-(aq) = fH of HCl (g) - SolutionH of HCl (g)
- fH of Cl-(aq) = -22.10 kcal mol-1 - (-17.9 kcal mol-1)
- fH of Cl-(aq) = -22.10 kcal mol-1 + 17.9 kcal mol-1
- fH of Cl-(aq) = -4.2 kcal mol-1
Answer:
- fH of Cl-(aq) = -4.2 kcal mol-1
Therefore, the correct answer is option 'A', which states that the fH of Cl-(aq) is -4.2 kcal mol-1.
Given Data:
- fH of HCl (g) = -22.10 kcal mol-1
- SolutionH of HCl (g) = -17.9 kcal mol-1
Calculating fH of Cl-(aq):
- The formation of Cl-(aq) from HCl (g) involves the dissociation of HCl into H+ and Cl- ions.
- The enthalpy change for this process is equal to the sum of the enthalpies of formation of the products minus the enthalpy of formation of the reactants.
- The enthalpy of formation of H+ is 0 kcal mol-1 as it is a free ion in solution.
- Therefore, fH of Cl-(aq) = fH of HCl (g) - SolutionH of HCl (g)
- fH of Cl-(aq) = -22.10 kcal mol-1 - (-17.9 kcal mol-1)
- fH of Cl-(aq) = -22.10 kcal mol-1 + 17.9 kcal mol-1
- fH of Cl-(aq) = -4.2 kcal mol-1
Answer:
- fH of Cl-(aq) = -4.2 kcal mol-1
Therefore, the correct answer is option 'A', which states that the fH of Cl-(aq) is -4.2 kcal mol-1.
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Given, ΔfH° of HCI (g) is - 22 . 10 kcal mol-1 and ΔSolutionH° (heat of solution) of HCI (g) is - 17.9 kcal mol-1. Thus, ΔfH ° of Cl-(aq) isa)-40.0 kcal mol-1b)+40.0 kcal mol-1c)+17.9 kcal mol-1d)+22.1 kcal mol-1Correct answer is option 'A'. Can you explain this answer?
Question Description
Given, ΔfH° of HCI (g) is - 22 . 10 kcal mol-1 and ΔSolutionH° (heat of solution) of HCI (g) is - 17.9 kcal mol-1. Thus, ΔfH ° of Cl-(aq) isa)-40.0 kcal mol-1b)+40.0 kcal mol-1c)+17.9 kcal mol-1d)+22.1 kcal mol-1Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Given, ΔfH° of HCI (g) is - 22 . 10 kcal mol-1 and ΔSolutionH° (heat of solution) of HCI (g) is - 17.9 kcal mol-1. Thus, ΔfH ° of Cl-(aq) isa)-40.0 kcal mol-1b)+40.0 kcal mol-1c)+17.9 kcal mol-1d)+22.1 kcal mol-1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Given, ΔfH° of HCI (g) is - 22 . 10 kcal mol-1 and ΔSolutionH° (heat of solution) of HCI (g) is - 17.9 kcal mol-1. Thus, ΔfH ° of Cl-(aq) isa)-40.0 kcal mol-1b)+40.0 kcal mol-1c)+17.9 kcal mol-1d)+22.1 kcal mol-1Correct answer is option 'A'. Can you explain this answer?.
Given, ΔfH° of HCI (g) is - 22 . 10 kcal mol-1 and ΔSolutionH° (heat of solution) of HCI (g) is - 17.9 kcal mol-1. Thus, ΔfH ° of Cl-(aq) isa)-40.0 kcal mol-1b)+40.0 kcal mol-1c)+17.9 kcal mol-1d)+22.1 kcal mol-1Correct answer is option 'A'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Given, ΔfH° of HCI (g) is - 22 . 10 kcal mol-1 and ΔSolutionH° (heat of solution) of HCI (g) is - 17.9 kcal mol-1. Thus, ΔfH ° of Cl-(aq) isa)-40.0 kcal mol-1b)+40.0 kcal mol-1c)+17.9 kcal mol-1d)+22.1 kcal mol-1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Given, ΔfH° of HCI (g) is - 22 . 10 kcal mol-1 and ΔSolutionH° (heat of solution) of HCI (g) is - 17.9 kcal mol-1. Thus, ΔfH ° of Cl-(aq) isa)-40.0 kcal mol-1b)+40.0 kcal mol-1c)+17.9 kcal mol-1d)+22.1 kcal mol-1Correct answer is option 'A'. Can you explain this answer?.
Solutions for Given, ΔfH° of HCI (g) is - 22 . 10 kcal mol-1 and ΔSolutionH° (heat of solution) of HCI (g) is - 17.9 kcal mol-1. Thus, ΔfH ° of Cl-(aq) isa)-40.0 kcal mol-1b)+40.0 kcal mol-1c)+17.9 kcal mol-1d)+22.1 kcal mol-1Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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Given, ΔfH° of HCI (g) is - 22 . 10 kcal mol-1 and ΔSolutionH° (heat of solution) of HCI (g) is - 17.9 kcal mol-1. Thus, ΔfH ° of Cl-(aq) isa)-40.0 kcal mol-1b)+40.0 kcal mol-1c)+17.9 kcal mol-1d)+22.1 kcal mol-1Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Given, ΔfH° of HCI (g) is - 22 . 10 kcal mol-1 and ΔSolutionH° (heat of solution) of HCI (g) is - 17.9 kcal mol-1. Thus, ΔfH ° of Cl-(aq) isa)-40.0 kcal mol-1b)+40.0 kcal mol-1c)+17.9 kcal mol-1d)+22.1 kcal mol-1Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Given, ΔfH° of HCI (g) is - 22 . 10 kcal mol-1 and ΔSolutionH° (heat of solution) of HCI (g) is - 17.9 kcal mol-1. Thus, ΔfH ° of Cl-(aq) isa)-40.0 kcal mol-1b)+40.0 kcal mol-1c)+17.9 kcal mol-1d)+22.1 kcal mol-1Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Given, ΔfH° of HCI (g) is - 22 . 10 kcal mol-1 and ΔSolutionH° (heat of solution) of HCI (g) is - 17.9 kcal mol-1. Thus, ΔfH ° of Cl-(aq) isa)-40.0 kcal mol-1b)+40.0 kcal mol-1c)+17.9 kcal mol-1d)+22.1 kcal mol-1Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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