Heat of formation of H₂O(g) at 25⁰C is -243 kJ,ΔE for the reaction

⇦

⇨

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.

Related Questions:

1. Which of the following undergoes sublimation
2. A drug that is antipyretic as well as analgesic is
3. The atomic number of cobalt is 27. The EAN of cobalt in Na₃[Co(NO₂)₄Cl₂] is
4. Which of the following is used as ”anasthesia”
5. The volume of 2.8g of carbon monoxide at 27⁰C and 0.821 atm pressure is

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