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Given, CH3COOH(aq) → CH3COO- (aq) + H+ (aq), ΔrH° = 0.005 kcal g-1
Enthalpy change when 1 mole of Ca(OH)2, a strong base, is completely neutralised by CH3COOH (aq) in dilute solution is
- a)-27.4 kcal mol-1
- b)-13.4 kcal mol-1
- c)-26.8 kcal mol-1
- d)-27.1 kcal mol-1
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Given, CH3COOH(aq) → CH3COO-(aq) + H+ (aq), ΔrH° = 0.0...
CH3COOH → CH3COO- + H+
∆H = 0.3 kcal mol-1
H+ + OH- → H2O
∆H = -13.7 kcal mol-1
Equation needed
2CCH3COH + Ca(OH)2 → 2CH3COO + 2H2O
∆H = 2×EQN(I) + 2×EQN(II)
∆H = (0.3)2 + (-13.7)2 = -26.8 kcal mol-1
∆H = 0.3 kcal mol-1
H+ + OH- → H2O
∆H = -13.7 kcal mol-1
Equation needed
2CCH3COH + Ca(OH)2 → 2CH3COO + 2H2O
∆H = 2×EQN(I) + 2×EQN(II)
∆H = (0.3)2 + (-13.7)2 = -26.8 kcal mol-1
Most Upvoted Answer
Given, CH3COOH(aq) → CH3COO-(aq) + H+ (aq), ΔrH° = 0.0...
To determine the enthalpy change when 1 mole of Ca(OH)2 is neutralized by CH3COOH, we can use the following equation:
Ca(OH)2(aq) + 2CH3COOH(aq) → Ca(CH3COO)2(aq) + 2H2O(l)
The enthalpy change for this reaction can be calculated using the enthalpy of formation values for the compounds involved. We can break down the reaction into smaller steps to calculate the overall enthalpy change.
Step 1: Dissociation of Ca(OH)2
Ca(OH)2(aq) → Ca2+(aq) + 2OH-(aq)
ΔH1 = ∑(ΔHf products) - ∑(ΔHf reactants)
The enthalpy change for the dissociation of Ca(OH)2 can be obtained from the enthalpy of formation values of the products and reactants involved. Since Ca2+ and OH- ions are in their standard states, their enthalpy of formation values are zero. Therefore, the enthalpy change for this step is zero.
Step 2: Formation of Ca(CH3COO)2
Ca2+(aq) + 2CH3COO-(aq) → Ca(CH3COO)2(aq)
ΔH2 = ∑(ΔHf products) - ∑(ΔHf reactants)
The enthalpy change for the formation of Ca(CH3COO)2 can be obtained from the enthalpy of formation values of the products and reactants involved. The enthalpy of formation of Ca(CH3COO)2 is not given in the question, but we can assume it to be zero since it is a strong base.
Step 3: Neutralization of CH3COOH
2CH3COOH(aq) + 2OH-(aq) → 2CH3COO-(aq) + 2H2O(l)
ΔH3 = ∑(ΔHf products) - ∑(ΔHf reactants)
The enthalpy change for the neutralization of CH3COOH can be obtained from the enthalpy of formation values of the products and reactants involved. The enthalpy of formation of CH3COO- and H2O are not given in the question, but we can assume them to be zero since they are in their standard states.
Overall Enthalpy Change:
ΔH = ΔH1 + ΔH2 + ΔH3
Since the enthalpy change for step 1 and step 2 are both zero, the overall enthalpy change is equal to the enthalpy change for step 3.
ΔH = ΔH3
Given that rH (the enthalpy change per gram) is 0.005 kcal g-1, we can calculate the enthalpy change per mole using the molar mass of Ca(OH)2.
Enthalpy change per mole = rH x molar mass
= 0.005 kcal g-1 x 74.1 g/mol (molar mass of Ca(OH)2)
= 0.3705 kcal mol-1
Therefore, the enthalpy change when 1 mole of Ca
Ca(OH)2(aq) + 2CH3COOH(aq) → Ca(CH3COO)2(aq) + 2H2O(l)
The enthalpy change for this reaction can be calculated using the enthalpy of formation values for the compounds involved. We can break down the reaction into smaller steps to calculate the overall enthalpy change.
Step 1: Dissociation of Ca(OH)2
Ca(OH)2(aq) → Ca2+(aq) + 2OH-(aq)
ΔH1 = ∑(ΔHf products) - ∑(ΔHf reactants)
The enthalpy change for the dissociation of Ca(OH)2 can be obtained from the enthalpy of formation values of the products and reactants involved. Since Ca2+ and OH- ions are in their standard states, their enthalpy of formation values are zero. Therefore, the enthalpy change for this step is zero.
Step 2: Formation of Ca(CH3COO)2
Ca2+(aq) + 2CH3COO-(aq) → Ca(CH3COO)2(aq)
ΔH2 = ∑(ΔHf products) - ∑(ΔHf reactants)
The enthalpy change for the formation of Ca(CH3COO)2 can be obtained from the enthalpy of formation values of the products and reactants involved. The enthalpy of formation of Ca(CH3COO)2 is not given in the question, but we can assume it to be zero since it is a strong base.
Step 3: Neutralization of CH3COOH
2CH3COOH(aq) + 2OH-(aq) → 2CH3COO-(aq) + 2H2O(l)
ΔH3 = ∑(ΔHf products) - ∑(ΔHf reactants)
The enthalpy change for the neutralization of CH3COOH can be obtained from the enthalpy of formation values of the products and reactants involved. The enthalpy of formation of CH3COO- and H2O are not given in the question, but we can assume them to be zero since they are in their standard states.
Overall Enthalpy Change:
ΔH = ΔH1 + ΔH2 + ΔH3
Since the enthalpy change for step 1 and step 2 are both zero, the overall enthalpy change is equal to the enthalpy change for step 3.
ΔH = ΔH3
Given that rH (the enthalpy change per gram) is 0.005 kcal g-1, we can calculate the enthalpy change per mole using the molar mass of Ca(OH)2.
Enthalpy change per mole = rH x molar mass
= 0.005 kcal g-1 x 74.1 g/mol (molar mass of Ca(OH)2)
= 0.3705 kcal mol-1
Therefore, the enthalpy change when 1 mole of Ca
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Given, CH3COOH(aq) → CH3COO-(aq) + H+ (aq), ΔrH° = 0.0...
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Given, CH3COOH(aq) → CH3COO-(aq) + H+ (aq), ΔrH° = 0.005 kcal g-1Enthalpy change when 1 mole of Ca(OH)2, a strong base, is completely neutralised by CH3COOH (aq) in dilute solution isa)-27.4 kcal mol-1b)-13.4kcal mol-1c)-26.8kcal mol-1d)-27.1kcal mol-1Correct answer is option 'C'. Can you explain this answer?
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Given, CH3COOH(aq) → CH3COO-(aq) + H+ (aq), ΔrH° = 0.005 kcal g-1Enthalpy change when 1 mole of Ca(OH)2, a strong base, is completely neutralised by CH3COOH (aq) in dilute solution isa)-27.4 kcal mol-1b)-13.4kcal mol-1c)-26.8kcal mol-1d)-27.1kcal mol-1Correct answer is option 'C'. 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, CH3COOH(aq) → CH3COO-(aq) + H+ (aq), ΔrH° = 0.005 kcal g-1Enthalpy change when 1 mole of Ca(OH)2, a strong base, is completely neutralised by CH3COOH (aq) in dilute solution isa)-27.4 kcal mol-1b)-13.4kcal mol-1c)-26.8kcal mol-1d)-27.1kcal mol-1Correct answer is option 'C'. 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, CH3COOH(aq) → CH3COO-(aq) + H+ (aq), ΔrH° = 0.005 kcal g-1Enthalpy change when 1 mole of Ca(OH)2, a strong base, is completely neutralised by CH3COOH (aq) in dilute solution isa)-27.4 kcal mol-1b)-13.4kcal mol-1c)-26.8kcal mol-1d)-27.1kcal mol-1Correct answer is option 'C'. Can you explain this answer?.
Given, CH3COOH(aq) → CH3COO-(aq) + H+ (aq), ΔrH° = 0.005 kcal g-1Enthalpy change when 1 mole of Ca(OH)2, a strong base, is completely neutralised by CH3COOH (aq) in dilute solution isa)-27.4 kcal mol-1b)-13.4kcal mol-1c)-26.8kcal mol-1d)-27.1kcal mol-1Correct answer is option 'C'. 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, CH3COOH(aq) → CH3COO-(aq) + H+ (aq), ΔrH° = 0.005 kcal g-1Enthalpy change when 1 mole of Ca(OH)2, a strong base, is completely neutralised by CH3COOH (aq) in dilute solution isa)-27.4 kcal mol-1b)-13.4kcal mol-1c)-26.8kcal mol-1d)-27.1kcal mol-1Correct answer is option 'C'. 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, CH3COOH(aq) → CH3COO-(aq) + H+ (aq), ΔrH° = 0.005 kcal g-1Enthalpy change when 1 mole of Ca(OH)2, a strong base, is completely neutralised by CH3COOH (aq) in dilute solution isa)-27.4 kcal mol-1b)-13.4kcal mol-1c)-26.8kcal mol-1d)-27.1kcal mol-1Correct answer is option 'C'. Can you explain this answer?.
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ample number of questions to practice Given, CH3COOH(aq) → CH3COO-(aq) + H+ (aq), ΔrH° = 0.005 kcal g-1Enthalpy change when 1 mole of Ca(OH)2, a strong base, is completely neutralised by CH3COOH (aq) in dilute solution isa)-27.4 kcal mol-1b)-13.4kcal mol-1c)-26.8kcal mol-1d)-27.1kcal mol-1Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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