Question

Chapter 9 59. The following sequence of reactions occurs in the commercial production of aqueous nitric acid: 4NH3(g)+502(g)-4NO(g)+6H2O(1)AH=-907kJ 2NO(g)+O2(g)—2NO2(g)AH--113kJ 3NO2+H2O(1)-2HNO3(aq)+NO(g)AH--139kJ Determine the total energy change for the production of one mole of aqueous nitric acid by this process.

Answer 1

59)

(1) --------------------       4 NH3(g) + 5O2(g) ---------------- 4 NO(g) + 6 H2O(l)     $\Delta$ H = - 907 KJ

(2) --------------------     2 NO(g) + O2(g) --------------------- 2 NO2(g)                     $\Delta$ H = - 113 KJ

(3) -------------------      3 NO2(g) + H2O(l) ------------------ 2 HNO3(aq) + NO(g) $\Delta$ H = - 139 KJ

The first equation is multiplied by 3

(1)   -------------         12 NH3(g) + 15 O2(g) ----------------12 NO(g) + 18 H2O(l)     $\Delta$ H = - 2721 KJ

The second equation is multipled by 6

(2) --------------       12 NO(g) + 6 O2(g) --------------------- 12 NO2(g)                     $\Delta$ H = - 678 KJ

The third equation is multiplied by 4

(3) ----------------- 12 NO2(g) + 4 H2O(l) ------------------ 8 HNO3(aq) + 4 NO(g) $\Delta$ H = - 556 KJ

The three equations are combined

                12 NH3(g) + 15 O2(g) ----------------12 NO(g) + 18 H2O(l)     $\Delta$ H = - 2721 KJ

              12 NO(g) + 6 O2(g) --------------------- 12 NO2(g)                     $\Delta$ H = - 678 KJ

              12 NO2(g) + 4 H2O(l) ------------------ 8 HNO3(aq) + 4 NO(g) $\Delta$ H = - 556 KJ

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       12 NH3(g) + 21 O2(g) ----------------- 8 HNO3(aq) + 4 NO(g) + 14 H2O(l)    $\Delta$ H = -2721 + (-678) + (-556)

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   12 NH3(g) + 21 O2(g) ----------------- 8 HNO3(aq) + 4 NO(g) + 14 H2O(l)        $\Delta$ H = - 3955 KJ

This is the energy change for the production of 8 moles of HNO3.

But, for one mole of production of aqueous solution og HNO3 = - 3955/8 = - 494.375

The energy change for the prouduction of one mole of aqueous solution of HNO3 = - 494.4 KJ