NAME: 2. Calculate q(heat) for the MgO/HCI reaction in units of kl. Calculate {

Question

NAME: 2. Calculate q(heat) for the MgO/HCI reaction in units of kl. Calculate { in units of kl/mole MgO. 3. Write the three thermochemical equations (1) to (3) (from pre-lab lecture) needed to values for each reaction (HINT: Look at calculate H for the reaction below. Include your answers for Q1 and Q2) 2Mg(s) + O2(g) 2MgO(s) 4. Neatly rewrite the three equations (multiply, reverse directions, etc.) such that when added, the result is the thermochemical equation above (4) for the combustion of magnesium. Calculate H for this reaction. (HINT: The thermochemical equations can be manipulated like algebraic equations)

Explanation / Answer

Ans. #1. Mass of solution = Volume of solution x Density

                                                = 100.0 mL x (1.02 g / mL)

                                                = 102.0 g

# Heat gained by a solution is given by-

q = m s dT                            - equation 1

Where,

q = heat gained

m = mass

s = specific heat

dT = Final temperature – Initial temperature

Putting the values in equation 1-

q = 102.0 g x (3.93 J g-1 0C-1) x 10.90C

Or, q = 4369.374 J

# The total amount of heat gained by the solution to increase its temperature must be equal to the total amount of heat released during the reaction between Mg and HCl.

So,

            Heat released during reaction between Mg and HCl = - 4369.374 J

Note: the –ve sign indicates that heat is being released during the reaction.

# Given, Mass of Mg = 0.244 g

Moles of Mg = Mass / Molar mass

                        = 0.244 g / (24.305 g/ mol)

                        = 0.0100 mol

# Now,

            dH = Amount of heat released / Moles of Mg reacted

            Or, dH = -4369.374 J / 0.0100 mol

            Or, dH = -436937.4 J/ mol

            Hence, dH = -436.94 kJ/ mol

Hence, dH for the reaction = -436.94 kJ/ mol

#2. Amount of heat released during MgO/HCl reaction is equal to the amount of heat gained by solution.

Or q = (101.0 mL x 1.02 g mol-1) x (3.93 J g-1 0C-1) x 7.40C

Or, q = 103.02 g x (3.93 J g-1 0C-1) x 7.40C

Hence, q = 2996.02764 J

Moles of MgO consumed = 0.996 g / (40.3044 g/ mol) = 0.024712 mol

Now,

            dH = - 2996.02764 J / 0.024712 mol = - 121237.76 J/mol = - 121.24 kJ/mol

#3. I. MgO(s) + 2 HCl(aq) ---------> MgCl2(aq) + H2O(l)           ; dH = -121.24 kJ/mol

      II. Mg(s) + 2 HCl(aq) ----------> MgCl2(aq) + H2(g)             ; dH = -436.94 kJ/mol

       III. H2(g) + ½ O2(g) ------------> H2O(l)                                ; dH = -285.8 kJ/mol

#4. When a reaction is reversed, the sign of dH is also reversed.

Following Hess’s Law, reaction IV can be written as sum of –

                        (Reverse of reaction I) + Reaction II + Reaction III

            MgCl2(aq) + H2O(l) ---------> MgO(s) + 2 HCl(aq)         ; dH = +121.24 kJ/mol

     (+) Mg(s) + 2 HCl(aq) ----------> MgCl2(aq) + H2(g)             ; dH = -436.94 kJ/mol         

     (+) H2(g) + ½ O2(g) ------------> H2O(l)                                  ; dH = -285.8 kJ/mol

Net:    Mg(s) + ½ O2(s) -----------> MgO(s)                             ; dHnet = ?      ; Rxn IV

dHnet for Rxn IV = (121.24 kJ/mol) + (-436.94 kJ/mol) + (-285.8 kJ/mol)

Hence, dHnet = -601.5 kJ/mol

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