Standard Enthalpy of Formation
Chemistry ⇒ Thermochemistry and Energetics
Standard Enthalpy of Formation starts at 11 and continues till grade 12.
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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}(g)] = -110 kJ/mol and \Delta H_f^\circ [\mathrm{CO_2}(g)] = -394 kJ/mol.
Describe how standard enthalpy of formation values are used to calculate the enthalpy change of a chemical reaction.
Describe the difference between standard enthalpy of formation and standard enthalpy of combustion.
Describe the importance of standard enthalpy of formation values in calculating enthalpy changes for chemical reactions.
Explain why the standard enthalpy of formation of \mathrm{Cl_2}(g) is zero.
Explain why the standard enthalpy of formation of diamond is not zero.
Explain why the standard enthalpy of formation of \mathrm{F_2}(g) is zero.
Explain why the standard enthalpy of formation of \mathrm{H_2O}(g) is different from that of \mathrm{H_2O}(l).
Given the following data: \Delta H_f^\circ [\mathrm{C_2H_5OH}(l)] = -277 kJ/mol, \Delta H_f^\circ [\mathrm{CO_2}(g)] = -394 kJ/mol, \Delta H_f^\circ [\mathrm{H_2O}(l)] = -286 kJ/mol. Calculate the enthalpy change for the reaction: \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{C_6H_{12}O_6}(s)] = -1273 kJ/mol, \Delta H_f^\circ [\mathrm{CO_2}(g)] = -394 kJ/mol, \Delta H_f^\circ [\mathrm{H_2O}(l)] = -286 kJ/mol. Calculate the enthalpy change for the reaction: \mathrm{C_6H_{12}O_6}(s) + 6 \mathrm{O_2}(g) \rightarrow 6 \mathrm{CO_2}(g) + 6 \mathrm{H_2O}(l).
Given the following data: \Delta H_f^\circ [\mathrm{CH_4}(g)] = -75 kJ/mol, \Delta H_f^\circ [\mathrm{CO_2}(g)] = -394 kJ/mol, \Delta H_f^\circ [\mathrm{H_2O}(l)] = -286 kJ/mol. Calculate the enthalpy change for the reaction: \mathrm{CH_4}(g) + 2 \mathrm{O_2}(g) \rightarrow \mathrm{CO_2}(g) + 2 \mathrm{H_2O}(l).
Given the following data: \Delta H_f^\circ [\mathrm{NaCl}(s)] = -411 kJ/mol, \Delta H_f^\circ [\mathrm{Na}(s)] = 0, \Delta H_f^\circ [\mathrm{Cl_2}(g)] = 0. Calculate the enthalpy change for the reaction: 2 \mathrm{Na}(s) + \mathrm{Cl_2}(g) \rightarrow 2 \mathrm{NaCl}(s).
Given the following data: \Delta H_f^\circ [\mathrm{NH_3}(g)] = -46 kJ/mol, \Delta H_f^\circ [\mathrm{N_2}(g)] = 0, \Delta H_f^\circ [\mathrm{H_2}(g)] = 0. Calculate the enthalpy change for the reaction: \mathrm{N_2}(g) + 3 \mathrm{H_2}(g) \rightarrow 2 \mathrm{NH_3}(g).
Given the following standard enthalpies of formation: \Delta H_f^\circ [\mathrm{NO_2}(g)] = 34 kJ/mol, \Delta H_f^\circ [\mathrm{NO}(g)] = 90 kJ/mol, calculate the enthalpy change for the reaction: 2 \mathrm{NO}(g) + \mathrm{O_2}(g) \rightarrow 2 \mathrm{NO_2}(g).
If the standard enthalpy of formation of \mathrm{NH_3}(g) is -46 kJ/mol, what does the negative sign indicate?
State Hess's Law and explain its relevance to standard enthalpy of formation.
State the standard conditions under which standard enthalpy of formation is measured.
The standard enthalpy of formation of \mathrm{C_2H_6}(g) is -84.7 kJ/mol. Write the formation equation for \mathrm{C_2H_6}(g).
Write the balanced chemical equation for the standard enthalpy of formation of \mathrm{CO_2}(g).
