Question

Calculate ΔHrxn for the following reaction:

CaO(s)+CO2(g)→CaCO3(s)

Use the following reactions and given ΔH values: Ca(s)+CO2(g)+12O2(g)→CaCO3(s), ΔH= -812.8 kJ

2Ca(s)+O2(g)→2CaO(s), ΔH= -1269.8 kJ

Express your answer using four significant figures.

Answer 1

Ans. Given,

Ca(s) + CO_2(g) + 1/2 O₂ --------> CaCO₃(s)           , dH = -812.8 kJ             - reaction 1

2 Ca(s) + O_2(g) --------> 2 CaO(s) , dH = -1269.8 kJ                       - reaction 2

CaO(s) + CO2(g) ----> CaCO3(s)           , dH = ?                        - reaction 3

Using Hess’s Law of constant heat summation, the enthalpy of a reaction is equal to the sum of two or more steps/reactions from which it derives.

Reaction 3 = (2 x reaction 1) + Reverse of reaction 2

            2 Ca(s) + 2 CO_2(g) + O₂ --------> 2 CaCO₃(s)         , dH = -1625.6 kJ         

        +   2 CaO(s) ------------> 2 Ca(s) + O_{2(g)                         ,}dH = + 1269.8 kJ

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            2 CaO(s) + 2 CO2(g) ----> 2 CaCO3(s)              , dH = 355.8 kJ

Note: 1. The sign of dH reversed if a reaction is written in reversed reaction.

2. The dH value of a reaction is multiplied with the same coefficient with which the reaction is multiplied.

So, we get,

2 CaO(s) + 2 CO2(g) ----> 2 CaCO3(s)                 , dH = - 355.8 kJ

Dividing above reaction with 2-

            CaO(s) + CO2(g) ----> CaCO3(s)                       , dH = - 177.9 kJ

Thus, dH_rxn = - 177.9 kJ