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Using the following thermochemical data, what is the change in enthalpy for the following reaction? Ca(OH)2(aq) + HCl(aq) CaCl2(aq) + 2H2O(l)CaO(s) + 2HCl(aq) CaCl2(aq) + H2O(l),  DH = -186 kJ CaO(s) + H2O(l) d Ca(OH)2(s), H = -65.1 kJ

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SvetkaChem
Ca(OH)2(aq) + 2HCl(aq)------> CaCl2(aq) + 2H2O(l) ΔH-?

CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l),  ΔH = -186 kJ

CaO(s) + H2O(l) -----> Ca(OH)2(s), ΔH = -65.1 kJ

1) Ca(OH)2 should be  reactant, so
CaO(s) + H2O(l) -----> Ca(OH)2(s) 
we are going to take as 
 Ca(OH)2(s)---->CaO(s) + H2O(l), and ΔH = 65.1 kJ

2) Add 2 following equations
Ca(OH)2(s)---->CaO(s) + H2O(l),                    and ΔH = 65.1 kJ
CaO(s) + 2HCl(aq)-----> CaCl2(aq) + H2O(l), and ΔH = -186 kJ

Ca(OH)2(s)+CaO(s) + 2HCl(aq)--->CaO(s) + H2O(l)+CaCl2(aq) + H2O(l)

Ca(OH)2(s)+ 2HCl(aq)---> H2O(l)+CaCl2(aq) + H2O(l)
By addig these 2 equation, we got the equation that we are needed,
so to find enthalpy of the reaction, we need to add  enthalpies of reactions we added.
ΔH=65.1 - 186 ≈ -121 kJ

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