Question

Given that H2(g) + F,() 2 HF(g) AH = -546.6 kJ 2 H,() + O2(g) → 2H,O(1) AH x = -571.6 kJ calculate the value of AHixn for 2F2(g) + 2 H2O(1) 4 HF(g) + O2(8) AH = KJ

Answer 1

law of thermodynamics, more one * According to hess's is reaction is obtained by adding than reaction then enthalpy charge of final reaction is obtained enthalpy change of above reactions. 9 B, AMI Bo > CAH2 by adding for e.g. А * if a AC ,AH = AHITAH₂ reaution is multiply by a nuneber then enthalpy change of final recution is also multiplied by the saure nouber- for e.g. A AHEX 2A -> 2B AH=27

for e.g. * if a scention is reversed, then enthalpy change of of reversed reaction is obtained by reversing the sign of enthalpy charge of original reaction A -> B, AH-X B-> A, AH=- Here, given (+F, (9) -> 2HF (१) if reaction is multiplied by 2, then en that by change of final reaction will be obtained by multi to enthalpy change of original reaction - 2H2 +2F2 UHE , OH 24-546.5)KJ H2 OH - 546.6 RJ multiplying 2 +2729) (9) 19) χη = 109 3:2 RJ 2H2 Given (9)+ 02 (3) > 2H20 (1), Aloxen -5716K reversing the above reaction, sign of an also revered, 2H2O(1) , AH -+57106RJ will be 242 9 toz 09) ran Now 2 H2 +2F2 UHE (9) (9) AH 1093.2K 2H20 (1) - 24209 +029) , a Hoxn: +571-6kg Adding the above two reautions, we get:

2H, оса) 2 Но +2+, е »чне. 25 (9) —чно, дно - о9 3.2 кО 259) О, 9 ) , Ана - S71-с К) +» чHE+e, у дню = -(093-2L) + 2 Ғ. 5+ 2H,о И ) (9) (9) + 71:6 — — Г2 (- 1к There, 2F2 (3) + 2H, о >чне +, 2 дни -S21- (K) (9) (9) (1) Atraen - - Г2 ( KJ