∆H,-572kJ/mol-1 for water It means that:
Correct answer: D. Water is at lower energy than its reactants
- A. It is an endothermic process
- B. Formation of water is a spontaneous process
- C. It is the heat evolved for complete reaction of 2 moles of oxygen with excess of hydrogen
- D. Water is at lower energy than its reactants
Explanation
The correct answer is that water is at lower energy than its reactants. This is supported by the negative enthalpy change (∆H = -572 kJ/mol), indicating that the reaction is exothermic, meaning energy is released when water is formed. This implies that the products (water) have lower potential energy compared to the reactants (hydrogen and oxygen).The other options are incorrect for the following reasons:The first option asserts that the process is endothermic, which contradicts the given negative enthalpy value.The second option discusses spontaneity, which is relevant but does not directly explain the relationship shown by the enthalpy change.The third option misinterprets the reaction stoichiometry; the enthalpy change refers to the formation of water, not specifically to the reaction of two moles of oxygen.
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.