Ca(s)+2H(aq)--> Ca^2+(aq)+H2 ∆H= -543 KJ/mol CaCO3(s)+2H(aq)-->Ca^2+(aq)+CO2(g)+H2O(l) ∆H= -15 KJ/mol Look up the heats formation for CO2...

Ca(s)+2H(aq)--> Ca^2+(aq)+H2 ∆H= -543 KJ/mol

CaCO3(s)+2H(aq)-->Ca^2+(aq)+CO2(g)+H2O(l) ∆H= -15 KJ/mol

Look up the heats formation for CO2 and H2O in the thermodynamic tables and use this information along with the above heats of reaction to calculate the standard enthalpy of formation for CaCO3. Write the balanced equations for the two heat of formations reactions, and use Hess Law in order to answer this question.

Expert Solution

C(s) + O2 (g) ------------> CO2(g) ∆H = -393.5Kj/mole

H2 + 1/2 O2(g) ----------> H2O (l) ∆H = -286 Kj/mole

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C(s) + H2(g)+ 3/2 O2(g) -------------> CO2(g) + H2O(l) ∆H = -679.5KJ/mole

C(s)+ H2(g) + 3/2 O2(g) -------------> CO2(g) + H2O(l) ∆H = -679.5KJ/mole

Ca(s)+2H(aq)--> Ca^2+(aq)+H2 ∆H= -543 KJ/mol

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Ca(s) + C(s) +2H(aq) + 3/2 O2(g) -------------> Ca^2+ (aq) + CO2(g) + H2O(l) ∆H = -1222.5 KJ/mole

CaCO3(s)+2H(aq)-- ----------------------------> Ca^2+(aq)+CO2(g)+H2O(l) ∆H= -15 KJ/mol

(-) (-) (-) (-) (-) (-) (+)

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Ca(s) + C(s) + 3/2 O2(g) --------------------> CaCO3(s) ∆H = -1207.5KJ/mole

queen_honey_blossom answered 2 years ago

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