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The enthalpy of neutralisation of HS- (aq) is - 5.1 kJ mol-1. Thus, second ionisation energy of H2S is
- a)+52.6 kJ mol-1
- b)-13.7 kJ mol-1
- c)-52.2 kJ mol-1
- d)+52.2 kJ mol-1
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
The enthalpy of neutralisation of HS-(aq) is - 5.1 kJ mol-1. Thus, sec...
∆Hneutralisation = ∆HH+ + OH- + ∆rHionisation
-5.1 = -57.3 + ∆Hionisation
∆Hionisation = 52.2 kJ mol-1
-5.1 = -57.3 + ∆Hionisation
∆Hionisation = 52.2 kJ mol-1
Most Upvoted Answer
The enthalpy of neutralisation of HS-(aq) is - 5.1 kJ mol-1. Thus, sec...
Second ionisation energy refers to the energy required to remove the second electron from an atom or ion. In this case, we are dealing with the second ionisation energy of H2S, which is the energy required to remove the second electron from the H2S molecule.
The enthalpy of neutralisation is the heat released when one mole of a substance is neutralized by a base. In this case, the enthalpy of neutralisation of HS- is given as -5.1 kJ mol-1. This means that when one mole of HS- reacts with a base, it releases 5.1 kJ of heat.
To determine the second ionisation energy of H2S, we can consider the neutralisation reaction between HS- and a base:
HS- + Base -> H2O + S2-
Since the enthalpy of neutralisation is negative, it indicates that the reaction is exothermic, meaning that heat is released during the reaction.
Now, let's focus on the second ionisation energy. The second ionisation energy of H2S can be calculated by considering the difference in enthalpy between the reactants and products of the neutralisation reaction:
H2S -> HS- + H+
The enthalpy change for this reaction is the negative of the enthalpy of neutralisation, which is 5.1 kJ mol-1.
Hence, the second ionisation energy of H2S is equal to the enthalpy change of the reaction:
Second Ionisation Energy = -5.1 kJ mol-1
Therefore, the correct answer is option 'D' (52.2 kJ mol-1).
The enthalpy of neutralisation is the heat released when one mole of a substance is neutralized by a base. In this case, the enthalpy of neutralisation of HS- is given as -5.1 kJ mol-1. This means that when one mole of HS- reacts with a base, it releases 5.1 kJ of heat.
To determine the second ionisation energy of H2S, we can consider the neutralisation reaction between HS- and a base:
HS- + Base -> H2O + S2-
Since the enthalpy of neutralisation is negative, it indicates that the reaction is exothermic, meaning that heat is released during the reaction.
Now, let's focus on the second ionisation energy. The second ionisation energy of H2S can be calculated by considering the difference in enthalpy between the reactants and products of the neutralisation reaction:
H2S -> HS- + H+
The enthalpy change for this reaction is the negative of the enthalpy of neutralisation, which is 5.1 kJ mol-1.
Hence, the second ionisation energy of H2S is equal to the enthalpy change of the reaction:
Second Ionisation Energy = -5.1 kJ mol-1
Therefore, the correct answer is option 'D' (52.2 kJ mol-1).
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The enthalpy of neutralisation of HS-(aq) is - 5.1 kJ mol-1. Thus, second ionisation energy of H2S isa)+52.6 kJ mol-1b)-13.7 kJ mol-1c)-52.2 kJ mol-1d)+52.2 kJ mol-1Correct answer is option 'D'. Can you explain this answer?
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The enthalpy of neutralisation of HS-(aq) is - 5.1 kJ mol-1. Thus, second ionisation energy of H2S isa)+52.6 kJ mol-1b)-13.7 kJ mol-1c)-52.2 kJ mol-1d)+52.2 kJ mol-1Correct answer is option 'D'. 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 The enthalpy of neutralisation of HS-(aq) is - 5.1 kJ mol-1. Thus, second ionisation energy of H2S isa)+52.6 kJ mol-1b)-13.7 kJ mol-1c)-52.2 kJ mol-1d)+52.2 kJ mol-1Correct answer is option 'D'. 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 The enthalpy of neutralisation of HS-(aq) is - 5.1 kJ mol-1. Thus, second ionisation energy of H2S isa)+52.6 kJ mol-1b)-13.7 kJ mol-1c)-52.2 kJ mol-1d)+52.2 kJ mol-1Correct answer is option 'D'. Can you explain this answer?.
The enthalpy of neutralisation of HS-(aq) is - 5.1 kJ mol-1. Thus, second ionisation energy of H2S isa)+52.6 kJ mol-1b)-13.7 kJ mol-1c)-52.2 kJ mol-1d)+52.2 kJ mol-1Correct answer is option 'D'. 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 The enthalpy of neutralisation of HS-(aq) is - 5.1 kJ mol-1. Thus, second ionisation energy of H2S isa)+52.6 kJ mol-1b)-13.7 kJ mol-1c)-52.2 kJ mol-1d)+52.2 kJ mol-1Correct answer is option 'D'. 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 The enthalpy of neutralisation of HS-(aq) is - 5.1 kJ mol-1. Thus, second ionisation energy of H2S isa)+52.6 kJ mol-1b)-13.7 kJ mol-1c)-52.2 kJ mol-1d)+52.2 kJ mol-1Correct answer is option 'D'. Can you explain this answer?.
Solutions for The enthalpy of neutralisation of HS-(aq) is - 5.1 kJ mol-1. Thus, second ionisation energy of H2S isa)+52.6 kJ mol-1b)-13.7 kJ mol-1c)-52.2 kJ mol-1d)+52.2 kJ mol-1Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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The enthalpy of neutralisation of HS-(aq) is - 5.1 kJ mol-1. Thus, second ionisation energy of H2S isa)+52.6 kJ mol-1b)-13.7 kJ mol-1c)-52.2 kJ mol-1d)+52.2 kJ mol-1Correct answer is option 'D'. Can you explain this answer?, a detailed solution for The enthalpy of neutralisation of HS-(aq) is - 5.1 kJ mol-1. Thus, second ionisation energy of H2S isa)+52.6 kJ mol-1b)-13.7 kJ mol-1c)-52.2 kJ mol-1d)+52.2 kJ mol-1Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of The enthalpy of neutralisation of HS-(aq) is - 5.1 kJ mol-1. Thus, second ionisation energy of H2S isa)+52.6 kJ mol-1b)-13.7 kJ mol-1c)-52.2 kJ mol-1d)+52.2 kJ mol-1Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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