Determine the enthalpy change for the decomposition of calcium carbonate CaCO_3

Question

Determine the enthalpy change for the decomposition of calcium carbonate CaCO_3 (s) rightarrow CaO(s) + CO_2(g) given the thermochemical equations below. Ca(OH)_2(s) rightarrow CaO (s) + H_2O(l) Delta_rH^o = 65.2 KJ/mol-rxn Ca(OH)_2(s) +CO_2 rightarrow CacO_3(s) +H_2O(l) Delta_rH^o = - 113.8 kJ/mol-rxn C(s) + O_2 (g) rightarrow CO_2 (g) Delta_rH^o = - 393.5 kJ/mol-rxn 2 Ca(s) + O_2 (g) rightarrow 2 Cao(s) Delta_rH^o = - 1270.2 kJ/mol-rxn +48.6 KJ/mol-rxn +179.0 KJ/mol-rxn +345.5 KJ/mol-rxn +441.0 KJ/mol-rxn +1711.7 kJ/mol-rxn Determine Delta_rH^o for the following reaction, 2 NH_3, (g) + 5/2 O_2(g) rightarrow 2 NO(g) + 3 H_2O(g) given the thermochemical equations below. N_2(g) + O_2 (g) rightarrow 2 NO(g) Delta_rH^o = +180.8 kJ/mol-rxn N_2(g) + 3H_2(g) rightarrow 2 NH_3(g) Delta_rH^o = -91.8 KJ/mol-rxn 2 H_2 (g) + O_2(g) rightarrow 2 H_2O(g)

Explanation / Answer

Revere equation-2 and add it to equation-1 we get,

CaCO3 + H2O + Ca(OH)2 --------> CO2 + CaO+H2O + CaOH)2

Here H2O and CA(OH)2 gets cancelled

Finally we get desired answer.

So enthalpy will be H1-H2

Delta H = 65.2-(-113.8) = 179 kJ/ mol

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