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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.
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Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- First law of thermodynamics is represented by the equation
- Which of the following is the smallest in size?
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- In the gas phase reaction C₂H₂ + H₂ ⇌ C₂H₆, the equilibrium constant
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Topics: Thermodynamics (179)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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