Which of the following is/are correct?I. When 1 mole of Zn is dissolved in excess HCl in an open beaker at 300 K and 1 atm, the work done is approximately - 2.49 kJ.II. When 1 mole of Zn is dissolved in excess HCl in a closed beaker, the work done is zero.
Correct answer: C. Both (I) and (II) are correct
- A. I only
- B. II only
- C. Both (I) and (II) are correct
- D. Neither (I) nor (II) are correct
Explanation
In the open beaker, H₂ gas is produced, and work is done by the system against the constant atmospheric pressure (W = -ΔngRT). In a closed, rigid beaker, the volume is constant, so no pressure-volume work is done. Both statements are correct.
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.