The standard enthalpy changes of formation of carbon dioxide and water are -394 KJmol and -286 kJ mol respectively. If the standard enthalpy change of combustion of propyne, CH3CCH is -1938 KJ mol-1 , what is its standard enthalpy change of formation?
Correct answer: C. -184 kJ mol-1
- A. +1258 kJ mol-1
- B. +180 kJ mol-1
- C. -184 kJ mol-1
- D. -680 kJ mol-1
Explanation
Option C is correct.The standard enthalpy change of formation of propyne, CH3CCH is -184 kJ/mol. The standard enthalpy change of formation is the heat released when a compound is formed from its constituent elements in their standard states. The standard enthalpy change of combustion is the heat released when a compound is burned in oxygen. The standard enthalpy change of combustion of propyne can be written as follows: CH3CCH(g) + 5O2(g) → 3CO2(g) + 4H2O(g) The standard enthalpy change of combustion of propyne is -1938 kJ/mol. This means that 1938 kJ of heat is released when 1 mole of propyne is burned in oxygen. The standard enthalpy changes of formation of carbon dioxide and water can be written as follows: CO2(g) → C(g) + O2(g) ΔHf° = -394 kJ/mol H2O(g) → H2(g) + ½O2(g) ΔHf° = -286 kJ/mol The standard enthalpy change of formation of propyne can be calculated using the following equation: ΔHf°(CH3CCH) = ΔHcomb°(CH3CCH) - ΔHf°(CO2) - 4ΔHf°(H2O) Substituting the values from the previous equations, we get: ΔHf°(CH3CCH) = -1938 kJ/mol - (-394 kJ/mol) - 4(-286 kJ/mol) ΔHf°(CH3CCH) = -184 kJ/mol Therefore, the standard enthalpy change of formation of propyne is -184 kJ/mol.
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.