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Heat of formation of H₂O(g) at 25⁰C is -243 kJ,ΔE for the reaction H₂(g) + 1/2 O₂(g) → H₂O(g) at 25⁰C is
Options
(a) -243 kJ
(b) -241.8 kJ
(c) 241.8 kJ
(d) 243 kJ
Correct Answer:
-241.8 kJ
Explanation:
H₂(g) + 1/2 O₂(g) → H₂O(g),
ΔHf = -243 kJ, ΔH = ΔE + Δn(g)RT,
where ΔH = enthalpy change of reaction = -243 kJ,
ΔE = internal energy change of reaction, Δn(g) = number of gaseous product – number of gaseous reactant = 1-(1+ 1/2) = – 1/2. ⇒ ΔE = ΔH – Δn(g)RT = -243000 + 0.5 x 8.314 x 298 = -241.76 kJ.
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Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which of the following sets of quantum numbers violates the rules of quantum mechanics
- Equal masses of H₂, O₂ and methane have been taken in a container of volume
- Biuret test is not given by
- At 298 K the enthalpy of formation of liquid water is -285.9 kJ/mol
- The products of the following reaction are
Topics: Thermodynamics (179)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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