The enthalpy of certain reactions at 273 K is -20.75 kJ. The enthalpy of the same reaction at 373 K (if heat capacities of reactants and products are the same) will be:
Correct answer: A. －20.75 kJ
- A. －20.75 kJ
- B. -2075 kJ
- C. Zero
- D. －20.75 x 373/273 kJ
Explanation
As it is provided in the given statement that heat capacities of reactants and products are the same, therefore there will be no energy change in the reaction. By Kirchoff's Rule; ∆Cp=∆H2-H1/T2-T1∆Cp(T2-T1)=∆H2-H1As, ∆Cp= 0 , ∆H2=H1 , therefore;H2= -20.75KJ
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.