| ⇦ | 
| ⇨ |
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.
Related Questions: - PV/T= Constant. A real gas will behave as ideal gas
- The structure of benzene is
- An aromatic compound among other things should have a bi-electron cloud containing
- An equation is given as:SnO₂ + C → Sn + CO.Its balanced form will involve
- The bakelite is prepared by the reaction between
Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- PV/T= Constant. A real gas will behave as ideal gas
- The structure of benzene is
- An aromatic compound among other things should have a bi-electron cloud containing
- An equation is given as:SnO₂ + C → Sn + CO.Its balanced form will involve
- The bakelite is prepared by the reaction between
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