Question

Determine the standard enthalpy of formation of calcium carbonate from the thermochemical equations given below.

Ca(OH)2(s) → CaO(s) + H2O()	ΔrH° = 65.2 kJ/mol-rxn
	Ca(OH)2(s) + CO2(g) → CaCO3(s) + H2O()	ΔrH° = −113.8 kJ/mol-rxn
	C(s) + O2(g) → CO2(g)	ΔrH° = −393.5 kJ/mol-rxn
	2 Ca(s) + O2(g) → 2 CaO(s)	ΔrH° = −1270.2 kJ/mol-rxn

a.

−1712.3 kJ/mol-rxn

b.

−1207.6 kJ/mol-rxn

c.

−980.6 kJ/mol-rxn

d.

−849.6 kJ/mol-rxn

e.

−441.8 kJ/mol-rxn

Answer 1

equation to be obtained : Ca(s) + C(s) + 3/2 O2 -> CaCO3

Ca(OH)2 +-> CaO + H2O H = 65.2 Kj/ mol - eqn 1

Ca(OH)2 + CO2 -> CaCO3 H = -113.8 KJ/mol - eqn 2

C + O2 -> CO2 H = -393.5KJ/mol - eqn 3

Ca + O2 -> CaO H = -1270.2KJ/ mol

adding eqn 4+ 2 * eqn 3 + 2 * eqn 2 - 2 * eqn 1

= -1270.2 + 2 * -393.5 + 2 * -113.8 - 2 * 65.2

=- 2415.2 KJ/mol

now , the above value is for the equation : 2Ca + 2C + 3O3 -> 2CaCO3

for 1 mole of CaCO3 , H = -2415.2/2 = -1207.6KJ/mol

option b is correct