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Standard Enthalpy of Formation

Chemistry ⇒ Thermochemistry and Energetics

Standard Enthalpy of Formation starts at 11 and continues till grade 12. QuestionsToday has an evolving set of questions to continuously challenge students so that their knowledge grows in Standard Enthalpy of Formation. How you perform is determined by your score and the time you take. When you play a quiz, your answers are evaluated in concept instead of actual words and definitions used.
See sample questions for grade 12
Calculate the standard enthalpy change for the reaction: 2CO(g) + O2(g) → 2CO2(g), given ΔHf°[CO2(g)] = -393.5 kJ mol-1 and ΔHf°[CO(g)] = -110.5 kJ mol-1.
Calculate the standard enthalpy change for the reaction: 2H2O(l) → 2H2(g) + O2(g), given ΔHf°[H2O(l)] = -285.8 kJ mol-1.
Describe how Hess's Law is used to determine the standard enthalpy of formation of a compound.
Describe the difference between standard enthalpy of formation and standard enthalpy of combustion.
Describe the significance of the sign (positive or negative) of the standard enthalpy of formation value.
Explain why the standard enthalpy of formation of a compound is important in thermochemistry.
Explain why the standard enthalpy of formation of C(diamond) is not zero.
Explain why the standard enthalpy of formation of noble gases is zero.
Explain why the standard enthalpy of formation of O3(g) is not zero.
Given the following data: ΔHf°[C2H5OH(l)] = -277.0 kJ mol-1, ΔHf°[CO2(g)] = -393.5 kJ mol-1, ΔHf°[H2O(l)] = -285.8 kJ mol-1, calculate the enthalpy change for the combustion of ethanol: C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l).
Given the following data: ΔHf°[CaO(s)] = -635.1 kJ mol-1, ΔHf°[CO2(g)] = -393.5 kJ mol-1, calculate the enthalpy change for the reaction: CaCO3(s) → CaO(s) + CO2(g), given ΔHf°[CaCO3(s)] = -1207.0 kJ mol-1.
Given the following data: ΔHf°[CO2(g)] = -393.5 kJ mol-1, ΔHf°[H2O(l)] = -285.8 kJ mol-1, calculate the enthalpy change for the combustion of methane: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l), given ΔHf°[CH4(g)] = -74.8 kJ mol-1.
Given the following data: ΔHf°[NH3(g)] = -46 kJ mol-1, ΔHf°[N2(g)] = 0, ΔHf°[H2(g)] = 0. What is the enthalpy change for the reaction: N2(g) + 3H2(g) → 2NH3(g)?
Given the following data: ΔHf°[NH4Cl(s)] = -314.4 kJ mol-1, ΔHf°[NH3(g)] = -46.0 kJ mol-1, ΔHf°[HCl(g)] = -92.3 kJ mol-1. Calculate the enthalpy change for the reaction: NH3(g) + HCl(g) → NH4Cl(s).
If the standard enthalpy of formation of a compound is highly negative, what does this indicate about the compound's stability?
State the standard enthalpy of formation of liquid water, H2O(l), given the following data: H2(g) + 1/2 O2(g) → H2O(l), ΔH = -285.8 kJ mol-1.
Write the standard enthalpy of formation equation for CaCO3(s).
A student claims that the standard enthalpy of formation of H2O2(l) is zero because it is composed of hydrogen and oxygen, both of which are elements. Critically evaluate this claim.
Consider the following reaction: C2H4(g) + H2(g) → C2H6(g). Given ΔHf°[C2H4(g)] = +52.3 kJ mol-1 and ΔHf°[C2H6(g)] = –84.7 kJ mol-1, calculate the standard enthalpy change for the reaction.
Explain why the standard enthalpy of formation of C(graphite) is zero, but that of C(diamond) is positive, even though both are forms of elemental carbon.