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: 2\mathrm{CO}(g) + \mathrm{O_2}(g) \rightarrow 2\mathrm{CO_2}(g), given \Delta H_f^\circ[\mathrm{CO_2}(g)] = -393.5 kJ mol^{-1} and \Delta H_f^\circ[\mathrm{CO}(g)] = -110.5 kJ mol^{-1}.

Calculate the standard enthalpy change for the reaction: 2\mathrm{H_2O}(l) \rightarrow 2\mathrm{H_2}(g) + \mathrm{O_2}(g), given \Delta H_f^\circ[\mathrm{H_2O}(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 \mathrm{O_3}(g) is not zero.

Given the following data: \Delta H_f^\circ[\mathrm{C_2H_5OH}(l)] = -277.0 kJ mol^{-1}, \Delta H_f^\circ[\mathrm{CO_2}(g)] = -393.5 kJ mol^{-1}, \Delta H_f^\circ[\mathrm{H_2O}(l)] = -285.8 kJ mol^{-1}, calculate the enthalpy change for the combustion of ethanol: \mathrm{C_2H_5OH}(l) + 3\mathrm{O_2}(g) \rightarrow 2\mathrm{CO_2}(g) + 3\mathrm{H_2O}(l).

Given the following data: \Delta H_f^\circ[\mathrm{CaO}(s)] = -635.1 kJ mol^{-1}, \Delta H_f^\circ[\mathrm{CO_2}(g)] = -393.5 kJ mol^{-1}, calculate the enthalpy change for the reaction: \mathrm{CaCO_3}(s) \rightarrow \mathrm{CaO}(s) + \mathrm{CO_2}(g), given \Delta H_f^\circ[\mathrm{CaCO_3}(s)] = -1207.0 kJ mol^{-1}.

Given the following data: \Delta H_f^\circ[\mathrm{CO_2}(g)] = -393.5 kJ mol^{-1}, \Delta H_f^\circ[\mathrm{H_2O}(l)] = -285.8 kJ mol^{-1}, calculate the enthalpy change for the combustion of methane: \mathrm{CH_4}(g) + 2\mathrm{O_2}(g) \rightarrow \mathrm{CO_2}(g) + 2\mathrm{H_2O}(l), given \Delta H_f^\circ[\mathrm{CH_4}(g)] = -74.8 kJ mol^{-1}.

Given the following data: \Delta H_f^\circ[\mathrm{NH_3}(g)] = -46 kJ mol^{-1}, \Delta H_f^\circ[\mathrm{N_2}(g)] = 0, \Delta H_f^\circ[\mathrm{H_2}(g)] = 0. What is the enthalpy change for the reaction: \mathrm{N_2}(g) + 3\mathrm{H_2}(g) \rightarrow 2\mathrm{NH_3}(g)?

Given the following data: \Delta H_f^\circ[\mathrm{NH_4Cl}(s)] = -314.4 kJ mol^{-1}, \Delta H_f^\circ[\mathrm{NH_3}(g)] = -46.0 kJ mol^{-1}, \Delta H_f^\circ[\mathrm{HCl}(g)] = -92.3 kJ mol^{-1}. Calculate the enthalpy change for the reaction: \mathrm{NH_3}(g) + \mathrm{HCl}(g) \rightarrow \mathrm{NH_4Cl}(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, \mathrm{H_2O}(l), given the following data: \mathrm{H_2}(g) + \frac{1}{2} \mathrm{O_2}(g) \rightarrow \mathrm{H_2O}(l), \Delta H = -285.8 kJ mol^{-1}.

Write the standard enthalpy of formation equation for \mathrm{CaCO_3}(s).

A student claims that the standard enthalpy of formation of \mathrm{H_2O_2}(l) is zero because it is composed of hydrogen and oxygen, both of which are elements. Critically evaluate this claim.

Consider the following reaction: \mathrm{C_2H_4}(g) + \mathrm{H_2}(g) \rightarrow \mathrm{C_2H_6}(g). Given \Delta H_f^\circ[\mathrm{C_2H_4}(g)] = +52.3 kJ mol^{-1} and \Delta H_f^\circ[\mathrm{C_2H_6}(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.