Question

4 Yanling He'sLaw write out Net knie Eqat on for reaction l and 2 and show ?11 for each a) b) Use reactions l and 2 and Hess' law to find the ?? for reaction 3, given here: c) Use the net ionic equation for reaction 3 to look up the AHe values in your texthook. Use sevaluestocalculation theoretical va elokupartheneath yourteethokUe these values to calculate a theoretical value for ??, or the "textbook" calc. 258

Answer 1

NaOH_aq + HCl_aq -------> H₂O_aq + NaCl_aq   

Reaction 1 : Net ionic equation : OH^-_(aq) + H⁺_(aq) -----------> H₂O_aq

Mg_(s) + 2HCl_aq -----> MgCl_2aq + H_2(g)

Reaction 2 : Net ionic equation : Mg_(s) + 2H⁺_(aq) ----------> Mg⁺² +  H_2(g)

$\Delta$ H_f of a reaction 1 : ? nH_products -   ? nH_reactants

(1 x $\Delta$ H_f (water) - ( 1 x $\Delta$ H_f OH^- + 1 x $\Delta$ H_f H⁺)

(1 x -285.83 ) - ( 1 x -229.9 + 1x 0 ) = -55.93 kJ/mole

$\Delta$ H_f of a reaction 1 : ? nH_products -   ? nH_reactants

(1 x $\Delta$ H_f (hydrogen) + 1 x $\Delta$ H_f (Mg⁺²)) - ( 1x $\Delta$ H_f (magnesium) + 1 x $\Delta$ H_f (H⁺)

( 1 x 0 + 1 X -466.85 ) - (1X 0 + 2 x 0) = -466.85 kJ/mol

Reaction 1 : Net ionic equation : OH^-_(aq) + H⁺_(aq) -----------> H₂O_aq $\Delta$ H_f = -55.93 kJ/mole

Reaction 2 : Net ionic equation : Mg_(s) + 2H⁺_(aq) ----------> Mg⁺² +  H_2(g) $\Delta$ H_f = - 466.85 kJ/mol

Reaction 3 : Mg_(s) + 2H₂0_(l) -------> Mg⁺²_(aq) + 2OH^-_(aq) + H_2(g)

Using Hess law , In order to obtain equation 3 reverse the reaction 1 and multiply by 2

Reverse Reaction 1 : Net ionic equation : ( H₂O_aq ----------->  OH^-_(aq) + H⁺_(aq) ) x 2    $\Delta$ H_f =(+55.93)2 kJ/mole

2H₂O_aq   -----------> 2OH^-_(aq) + 2H⁺_(aq) $\Delta$ H_f = +111.86 kJ/mole

Reaction 2 : Net ionic equation : Mg_(s) + 2H⁺_(aq) ----------> Mg⁺² +  H_2(g) $\Delta$ H_f = - 466.85 kJ/mol

Reaction 3 : Mg_(s) + 2H₂0_(l) -------> Mg⁺²_(aq) + 2OH^-_(aq) + H_2(g) $\Delta$ H_f = -466.85 +111.86 = -354.99 kJ/mol

Theortical value of $\Delta$ H_f of reaction 3 is

$\Delta$ H_f of a reaction 3 : ? nH_products -   ? nH_reactants

( 1 x $\Delta$ H_f of Mg⁺²_(aq) + 2 x $\Delta$ H_f of OH^- + 1 X $\Delta$ H_f of H₂) - (1 X $\Delta$ H_f of Mg + 2 x $\Delta$ H_f of water)

( 1 x -466.85 + 2 x -229.9 + 1 x0) - ( 1 x 0 + 2 x -285.83)

(-926.65)-(-571.66) = -354.99 kJ/mol