You measure the following heats of reaction at 298 K Ca (s) + 2H + (aq) ------>

Question

You measure the following heats of reaction at 298 K Ca (s) + 2H+ (aq) ------> Ca2+ (aq) +H2(g)                                 H= -543 Kj/mol CaCO3 (s) + 2H+ (aq) ----->Ca2+ (aq) + CO2 (g) + H2O(l)      H= -15kj/mol From thermodynamic tables, you find f CO2 (g) = -394 KH/mol, and f(H2O, l )= -286 kj/mol at 298 K. Calculatef (CaCO3, s) at 298 k. You measure the following heats of reaction at 298 K Ca (s) + 2H+ (aq) ------> Ca2+ (aq) +H2(g)                                 H= -543 Kj/mol CaCO3 (s) + 2H+ (aq) ----->Ca2+ (aq) + CO2 (g) + H2O(l)      H= -15kj/mol From thermodynamic tables, you find f CO2 (g) = -394 KH/mol, and f(H2O, l )= -286 kj/mol at 298 K. Calculatef (CaCO3, s) at 298 k.

Explanation / Answer

The best way to do this is to add the equations and thierenergies like an addition problem like this If you reverse areaction don't forget to reverse the sign of its energy. Alsodont' forget to rate the answers you recieve. Heat of formation isthe heat to form it from one mole of the product from itselements in thier base form. For CaCO3 this wouldbe Ca (s), C(s), and O2(g). Ca2+ (aq) + CO2 (g) + H2O (l) ----->  CaCO3 (s)+ 2 H+ (aq)      15kJ/mol Ca (s) + 2H+ (aq) ------> Ca2+ (aq) +H2(g)                                 -543 kJ/mol C (s) + O2(g) -------> CO2(g)                                                   -394 kJ/mol H2(g) + 1/2 O2(g) -------->H2O(l)                                             -286 kJ/mol ---------------------------------------------------------------------------------- Ca(s) + C (s) + 1.5 O2(g) ------>CaCO3(s)                              -1208 kJ/mol

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