Consider the reaction Solution For Part A: Standard Enthalpy of formation ______

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Consider the reaction

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For Part A: Standard Enthalpy of formation ______________________________________ NaOH (s) = -425.93 kJ/mol ; CO2 (g) = -393.509 kJ/mol ; Na2CO3 (s) = -1130.77 kJ/mol ; H2O (l) = -285.8 kJ/mol ; for the 1st reaction 2 NaOH(s) + CO2 (g) ----> Na2CO3 (s) + H2O (l) Enthalpy of reation = Enthalpy products minus Enthalpy reactants = (-1130.77 + -285.8) - ( -425.93*2 + -393.509 ) = ( -1416.57 ) - ( -1245.369 ) = -171.201 kJ/mol. **************************************************************************************** For Part B: let the Enthalpy of Na2O (s) = x kJ/mol. Enthalpy of products { Na2O (s) + CO2 (g) } minus Enthalpy of Reactants { Na2CO3 (s) } = 321.5 kJ/mol ------>>> ( x + -393.509) - ( -1130.77 ) = 321.5 ------->>>> x -393.509 + 1130.77= 321.5 ----->>> x=321.5+393.509 - 1130.77 = -415.761 kJ/mol.

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