For a reaction 2 NH3 ➞ N2,+ 3H2, which of the following statements is correct.
Correct answer: C. ∆H = ∆E
- A. ∆H = 0
- B. ∆H > ∆E
- C. ∆H = ∆E
- D. ∆H < ∆E
Explanation
For the reaction 2 NH3 ➞ N2 + 3 H2, the correct statement is that ∆H = ∆E. This is because, in many reactions at constant pressure, the change in enthalpy (∆H) and the change in internal energy (∆E) can be equal, particularly when there is no significant pressure-volume work involved. In this decomposition reaction, the work done on or by the system is minimal, leading to ∆H equaling ∆E. Other options are incorrect because they either misrepresent the relationship between ∆H and ∆E or wrongly assume ∆H as zero, which is not characteristic of an exothermic reaction.
Last updated
About Thermochemistry and Energetics of Chemical Reactions
Thermochemistry measures energy changes in chemical reactions and distinguishes exothermic from endothermic processes. Work covers systems, surroundings and state functions, internal energy, the first law of thermodynamics, enthalpy and Hess's law, including the sign conventions used when heat enters or leaves a system.
Practise Thermochemistry and Energetics of Chemical Reactions
728 free Thermochemistry and Energetics of Chemical Reactions MCQs from Chemistry, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Chemistry questions like this
Chemistry is on 12 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
1/2 H2(g) ➞ H(g) ∆H = 218 kJ mol-1.In this reaction, ∆H will be called:
1 Kcal is equal to:
2P(s) 3CI2(g) 2PCI3 ∆H= 151.8 kJ PCl3+ CI2(g) , PCI5(g); ∆H = -32.8 kJFrom the following data, the heat of formation of PCI5 comes out to be :
5 calories are equivalent to _ Joule.
5 mole of an ideal gas expand reversibly from a volume of 8 dm3 to 80 dm3 at a temperature of 27°C. Calculate the change in entropy.