Question

1. Given the following equations and H^o values given below, determine the heat of reaction at 298 K for the reaction:

2 N₂(g) + 5 O₂(g) 2 N₂O₅(g)

2 H2(g) + O2(g)  2 H2O(l)                   Ho/kJ = -571.6

N2O5(g) + H2O(l)  2 HNO3(l)             Ho/kJ = -73.7

N2(g) + 3 O2(g) + H2(g)  2 HNO3(l) Ho/kJ = -348.2

Answer 1

The heat of the reaction at 298 K for the given reaction is +22.6 kJ .

The detail calculation is shown below .

the We have to calculate heat of the reaction at 2988 (40) for the reeaction : 2N, (g) 502 (9) - 2N205 g) 2 H₂ (g) + O₂(g) 2H₂0 (1); 4" - 571.6KJ. N205 (g) Holl) = 2 HNO 3 (1); H = -73.7 kJ N₂ (g) 30₂(g) + H₂(g) = 2 HNO3(0);">-348-2 KJ [2x equation (on)]- Lequation (i)), we get 2 N₂ (9) + 602 (9) + 2H₂(g) = 4HNO3 (1). (-348.2x2) kJ -- 696.4 kJ. 2H₂ (9) & 0₂ (9) = 2H₂0 (1); A = -571.6 kJ 2 N₂(g) + 602(g) + 24219) + 24₂0 (1) 4HNO3 (1) + 243 (9) + 02(9). 2N₂(g) +502 (9) + 2H₂0 (1) AHNO3(e) (in A4°=5696.4-(-541.65] «T - 124.8KJ.

[equation liv] - (2x equations (it)], we get 2N, (g) +509 (8) + 2H20 0) = 4HNO3(e); H'= -124.sk] 2N₂05 (9) + 2 H₂0 (1) 4 4N03 (); 4°= (-73.442)KJ -- 147.4 KJ 2N, (0) +502(1) + 2.43660) + 4 uNog (4) 4HNO3 () + 2 N205 (9) + 24 2N₂ (9) +502 (9) = 2N₂05 (9) **--124.9-(-147.43] 22.6K J the value of heat of reaction (tre) at 2988 +22.6 KI.