The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps:4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)(1)2NO(g) + O2(g) → 2NO2(g)(2)3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g)(3)Use the values of ΔH°f to calculate the value of ΔH° for each of the preceding reactions.ΔH°f, kJ/molNH3(g)-46NO(g)90.H2O(g)-242NO 2(g)34H2O(l)-286HNO3(aq)-207ΔH°1 = kJΔH°2 = kJΔH°3 = kJWrite the overall equation for the production of nitric acid by the Ostwald process by combining the preceding equations. (Water is also a product.)(Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.)

ΔH°1 = -908 kJ

ΔH°2 = -112 kJ

ΔH°3 = -140 kJ

Combination of the reactions for Ostwald Process' overall reaction

4NH₃ + 8O₂ → 4HNO₃ + 4H₂O

Total Heat of reaction = -1524 kJ

On a per mole basis

NH₃ + 2O₂ → HNO₃ + H₂O

Total Heat of reaction = -381 kJ/mol

Explanation:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

2NO(g) + O₂(g) → 2NO₂(g)

3NO₂(g) + H₂O(l) → 2HNO₃(aq) + NO(g)

ΔH°f, kJ/mol

NH3(g), -46

NO(g), 90.

H2O(g), -242

NO2(g), 34

H2O(l), -286

HNO3(aq), -207

We are to find the heat of reaction for the 3 leading reactions. ΔH°1, ΔH°2, ΔH°3

The heat of any reaction is given as the

ΔH(products) - ΔH(reactants)

For reaction 1, ΔH°1

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

ΔH°1 = ΔH(products) - ΔH(reactants)

ΔH(products) = (4×90) + (6×-242) = -1092 KJ

ΔH(reactants) = (4×-46) + (5×0) = -184 KJ

ΔH°1 = -1092 - (-184) = -908 KJ

For reaction 2, ΔH°2

2NO(g) + O₂(g) → 2NO₂(g)

ΔH°2 = ΔH(products) - ΔH(reactants)

ΔH(products) = (2×34) = 68 kJ

ΔH(reactants) = (2×90) + (1×0) = 180 kJ

ΔH°2 = 68 - 180 = -112 kJ

For reaction 3, ΔH°3

3NO2(g) + H₂O(l) → 2HNO₃(aq) + NO(g)

ΔH°3 = ΔH(products) - ΔH(reactants)

ΔH(products) = (2×-207) + (1×90) = -324 kJ

ΔH(reactants) = (3×34) + (1×-286) = -184 KJ

ΔH°3 = -324 - (-184) = -140 kJ

For the combination of the reactions,

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

2NO(g) + O₂(g) → 2NO₂(g) ×3

6NO(g) + 3O₂(g) → 6NO₂(g)

3NO₂(g) + H₂O(l) → 2HNO₃(aq) + NO(g) ×2

6NO₂(g) + 2H₂O(l) → 4HNO₃(aq) + 2NO(g)

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

6NO(g) + 3O₂(g) → 6NO₂(g)

6NO₂(g) + 2H₂O(l) → 4HNO₃(aq) + 2NO(g)

Summing it all up, and cancelling the species that appear on both sides.

4NH₃ + 8O₂ → 4HNO₃ + 4H₂O

Total ΔH°f according to Born-Haber cycle and Hess' law.

Total ΔH°f = ΔH°1 + 3ΔH°2 + 2ΔH°3

= -908 + (3×-112) + (2×-140)

= -1524 kJ

On a per mole basis

ΔH°f = (-1524 ÷ 4) = -381 kJ/mol

Hope this Helps!!!

nacl is added to the aqueous of what? ? in order to get a proper answer you have to specify the exact composition of the aqueous solution.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml