| ⇦ | 
| ⇨ |
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: - Which of the following statements is correct regarding the drawbacks of raw rub
- If the value of Cp for nitrogen gas is 7 JK⁻¹mol⁻¹ , then the value
- Which of the following element,has the same oxidation state in all of its com.
- The lattice energy order for lithium halide is
- Chinese white is
Topics: Thermodynamics
(179)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which of the following statements is correct regarding the drawbacks of raw rub
- If the value of Cp for nitrogen gas is 7 JK⁻¹mol⁻¹ , then the value
- Which of the following element,has the same oxidation state in all of its com.
- The lattice energy order for lithium halide is
- Chinese white 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
Leave a Reply