Question

The standard enthalpy of formation (ΔHf°) of calcium carbonate is –1207 kJ/mol. Which ONE of the equations below has ΔH° = –1207 kJ? A) Ca(s) + C(s) + 3/2 O2(g) –––––> CaCO3(s) B) 2 Ca(s) + 2 C(s) + 3 O2(g) –––––> 2 CaCO3(s) C) Ca(s) + C(s) + 3 O(g) –––––> CaCO3(s) D) CaCO3(s) ––––> Ca(s) + C(s) + 3/2 O2(g) E) CaO(s) + CO2(g) –––––> CaCO3(s)

Answer 1

Answer:

A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)

Explanation:

Standard enthalpy of formation of a chemical is defined as the change in enthalpy durin the formation of 1 mole of the substance from its constituent elements in their standard states.

The consituent elements of calcium carbonate, CaCO₃, in their standard states (States you will find this pure elements in nature), are:

Ca(s), C(s) and O₂(g)

That means, the equation that represents standard enthalpy of CaCO₃ is:

A) Ca(s) + C(s) + 3/2 O₂(g) → CaCO₃(s)

Is the equation that has ΔH° = -1207kJ/mol