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Heat of combustion ΔH for C(s) , H₂(g)and CH₄(g) are -94, -68 and -213 kcal/mol. Then ΔH for C(s) + 2H₂(g) → CH₄(g)is
Options
(a) -17 kcal
(b) -111kcal
(c) -170 kcal
(d) -85 kcal
Correct Answer:
-17 kcal
Explanation:
(i) C(s) + O₂ = CO₂ ; ΔH(i) = -94 kcal/mol,
(ii) 2H₂ + O₂ = 2H₂O ; ΔH(ii) = -68 x 2 kcal/mol,
(iii) CH₄ + 2O₂ = CO₂ + 2H₂O ; ΔH(iii) = -213 kcal/mol,
(iv) C(s) + 2H₂(g) = CH₄(g) ; ΔH(iv) = ?,
By applying Hess’s law we can compute ΔH(iv), (i) + (ii) – (iii), we have
C + O₂ + 2H₂ + O₂ – CH₄ – 2O₂ = CO₂ + 2H₂ O – CO₂ – 2H₂ O,
C+ 2H₂ = CH₄,
Therefore ΔH(iv) = ΔH(i) + ΔH(ii) – ΔH(iii) = (-94 – 68 x 2 + 213 ) kcal = -17 kcal.
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Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The anion, (Si₆O₁₈)¹²⁻ is present in
- In the conversion of Br₂ to BrO₃⁻, the oxidation number of Br changes from
- Which of the following species contains three bond pairs and one lone pair
- If concentration of reactants is increased by ‘x’,then the K becomes
- In Kjeldahl’s method,ammonia from 5g of food neutralizes 30 cm3 of 0.1 N acid.
Topics: Thermodynamics (179)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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