⇦ | ⇨ |
The standard heat of formation of carbon disulphide (l) given that the standard heat of combustion of carbon(s) , sulphur(s) and carbon disulphide(l) are -393.3, -293.72 and -1108.76 kJ mol⁻¹ respectively is
Options
(a) 128.02 kJ mol⁻¹
(b) 12.802 kJ mol⁻¹
(c) -128.02 kJ mol⁻¹
(d) -12.802 kJ mol⁻¹
Correct Answer:
128.02 kJ mol⁻¹
Explanation:
(i) C(s) + O₂(g) → CO₂(g); ΔH₁ = -393.3 kJ.
(ii) S(s) + O₂(g) → SO₂(g); ΔH₂ = -293.72 kJ.
(iii) CS₂(l) + 3O₂(g) → CO₂ + 2SO₂ ; ΔH₃ = -1108.76 kJ.
On adding (i) and (ii) and subtracting (iii), we get,
C(s) + 2S(s) → CS₂(g), ΔH = -393.3 + 2 (-293.72) + 1108.76 = +128.02 kJ mol⁻¹.
Related Questions: - Which of the following organic compounds polymerizes to form the polyester Dacron
- Reduction of H₃C – NC with hydrogen in presence of Ni/Pt as catalyst gives
- Glucose and mannose are
- Which one of the following compounds will react with NaHCO₃ solution to give
- If the gas at constant temperature and pressure expands, then its
Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which of the following organic compounds polymerizes to form the polyester Dacron
- Reduction of H₃C – NC with hydrogen in presence of Ni/Pt as catalyst gives
- Glucose and mannose are
- Which one of the following compounds will react with NaHCO₃ solution to give
- If the gas at constant temperature and pressure expands, then its
Topics: Thermodynamics (179)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply