Record your values for HA and HB below: Mg (s) + 2 HCl (aq) MgCl2 (aq) + H2 (g)

Question

Record your values for HA and HB 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 = ____________________

Explanation / Answer

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

----------------------------------------------------------------------------------------------------------------------------

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

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