Question

Nitric acid can be formed in two steps from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming natural gas. In the first step, nitrogen and hydrogen react to form ammonia: N (0) + 3H (9) 2 NH,(9) AH = -92. kJ In the second step, ammonia and oxygen react to form nitric acid and water: NH (g) + 20 (g) – INO,g) + H20() AH -330, RJ Calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these reactions. Round your answer to the nearest kJ. $ ?

Answer 1

Given: N2(g) + 3H₂(g) 2NH3; AH, = -92 kJ -0 NH3 (9) + 20, (d) — HNO3(g) + 4.009); $4=330 kJ © Target escolation : _ +3H2 +20 - HNO3 + H20 ; AH= ? | We, know, thermo chemical equation can be treated as mathematical еслаtом Therefore, applying the operation : 4 x 0 + That is, I Nz (g) + 3 (g) - NH3(g); $H, = -92 kJ + NHolg) + 2Q (g) HNO3 (g) + H₂O(g) AH = -330k] ££ Nz (g) + 3 2 (g) +2ą - HNO3(g) + 20 (g) ; AH = AH, HAL Target Equation AH = AH, + Olta 41-(2-330) ko [ 41= -31667