Question

Calculate the standard enthalpy change for the following reaction at 298 K: 2NaHCO3(s) + MgCl2(aq) → MgCO3(s) + 2NaCl(aq) + H2O(l)Given the following standard enthalpies of formation: ΔHf°(NaHCO3) = -950.7 kJ/molΔHf°(MgCl2) = -641.8 kJ/molΔHf°(MgCO3) = -1128.2 kJ/molΔHf°(NaCl) = -411.2 kJ/molΔHf°(H2O) = -285.8 kJ/mol

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following equation:H reaction  =  [Hf products ] -  [Hf reactants ]For the reactants:2 moles of NaHCO3: 2 *  -950.7 kJ/mol  = -1901.4 kJ1 mole of MgCl2: 1 *  -641.8 kJ/mol  = -641.8 kJFor the products:1 mole of MgCO3: 1 *  -1128.2 kJ/mol  = -1128.2 kJ2 moles of NaCl: 2 *  -411.2 kJ/mol  = -822.4 kJ1 mole of H2O: 1 *  -285.8 kJ/mol  = -285.8 kJNow, we can plug these values into the equation:H reaction  = [ -1128.2  +  -822.4  +  -285.8 ] - [ -1901.4  +  -641.8 ]H reaction  =  -2236.4  -  -2543.2 H reaction  = 306.8 kJThe standard enthalpy change for the reaction at 298 K is 306.8 kJ.