Question

Record your values for ΔHA and ΔHB below: Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g) ∆HA = __-4.70 x 105 kJ/mol Mg_ MgO (s) + 2 HCl (aq) → MgCl2 (aq) + H2O (l) ∆HB = __-150 kJ/mol MgO___ _ In order to calculate the heat of formation of MgO (∆Hf MgO), you will also need one additional enthalpy value, the enthalpy of formation of liquid water. Look this value up in a textbook and record it below. Be sure you record the value for liquid water, not gaseous water. H2 (g) + ½ O2 (g) → H2O (l) ∆Hf = _-285.8 kJ/mol__ Calculations: Show all of your work, with units. Record answers with correct significant figures. 1. Using the information determined above and Hess’s Law, the heat of formation (∆Hf) for MgO can be obtained. Rearrange the three equations above to determine H for: Mg (s) + ½ O2 (g)  MgO (s): show your work, and pay attention to sig figs! Mg (s) + ½ O2 (g) → MgO (s) H2 (g) + ½ O2 (g) → H2O (l) Your value for ∆Hf MgO(pay attention to sig figs!) = __________________ 2. Look up the theoretical (textbook) value for the heat of formation of MgO and calculate your percent difference to the correct significant figures. Textbook value for ∆Hf MgO = __-601.6__ kJ/mol % difference = ____________________

Answer 1

Mg (s) + 2 HCl (aq) --------------> MgCl2 (aq) + H2 (g) , ∆HA = -4.70 x 105 kJ/mol ------------->1

MgO (s) + 2 HCl (aq) ------------> MgCl2 (aq) + H2O (l) ,  ∆HB = -150 kJ/mol ------------------->2

MgCl2 (aq) + H2 (g) ------------>  Mg (s) + 2 HCl (aq) ,   ∆HB = + 4.70 x 105 kJ/mol ---------------->3 (reverse of 1)

add 2 +3

MgO (s) + 2 HCl (aq) ------------> MgCl2 (aq) + H2O (l) ,  ∆H = -150 kJ/mol

MgCl2 (aq) + H2 (g) ------------>  Mg (s) + 2 HCl (aq) ,   ∆H = + 4.70 x 10^2 kJ/mol

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MgO + H2 --------------------------> Mg + H2O ,   ∆Hrxn = 320 kJ/mol

∆Hrxn =  ∆H products -  ∆H reactants

320 = (-285.8) - ∆HMgO

∆HMgO = - 605.8 kJ / mol

% difference = (605.8 - 601.6 / 605.8 ) x 100

= 0.70 %

note : you have given delta HA wrong value . it should not be . check once