UIVLII Lidl H, (g)+F2(g) → 2 HF(g) 2 H2(g) + O2(g) — 2 H2O(1) AH;xn = -546.6 kJ AHixn = -571.6 kJ calculate the value o...
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
Given that H2(g) + F2(g) → 2 HF(g) AH;xn = -546.6 kJ 2 H2(g) + O2(g) + 2H,0(1) AH;x= -571.6 kJ calculate the value of AHEX for 2F2(g) + 2 H20(1) + 4HF(g) + O2(8) AH
18. Given that AHorxn = -546.6 kJ H2 (g)+F2 (g)2HF (g) AH°rxn = -571.6 kJ 2H2 (g)+ O2 (g) 2H20 (1) Calculate the value of AHrxn for 2F2 (g) 4HF (g) + O2 (g) +2H20 (I)
Given that H2(g) +F2(g) → 2HF(g) 2 H, (9) +0,(8) — 2H,O(1) calculate the value of AHin for A Hisn = -546.6 kJ mol- A Hixn = -571.6 kJ mol- 2F2(g) + 2 H,O(1) — 4 HF(g) + O2(g) AH = kJ.mol-1 KJ-mol-'
H ) + F(g) --- 2 HF(g) A Hin = -546.6 kJ 2.H. (9) +0,() -2H,O(1) AH = -571.6 kJ calculate the value of A His for 2F,(g) + 2 H, 0(1) --- 4 HF(g) + O2(g) J K
Given that AHn-546.6 kJ H,(g)+F,(g)2 HF(g) 2H, (g)+0,(g)2H,O() AH -571.6 kJ calculate the value of AHn for 4 HF(g)+O,(g) 2F,(g)+2H,O0) AH= kJ
Given that AHn-546.6 kJ H,(g)+F,(g)2 HF(g) 2H, (g)+0,(g)2H,O() AH -571.6 kJ calculate the value of AHn for 4 HF(g)+O,(g) 2F,(g)+2H,O0) AH= kJ
Given that H,() +F,() 2 HF(8) 2 H,(8) + O2(8) + 2H,0(1) AH:n = -546.6 kJ AHan = -571.6 kJ calculate the value of AHin for 2F%(8) + 2H,0(1) 4 HF(g) + O2(E) AH
2. Use Hess’s Law to determine the enthalpy of the reaction below. 2F2(g) + 2H2O(l) → 4HF(aq) + O2(g) DH˚= ? H2(g) + F2(g) → 2HF(aq) DH˚ = -546.6 kJ 2H2 (g) + O2(g) → 2H2O(l) DH˚ = -571.6 kJ a. 42 kJ b. -1120 kJ c. -251 kJ d. -521 kJ e. -1690 kJ
5. Given the following data: 2 H2(g) + O2(g) → 2 H20 (1) AH° = -571.6 kJ N,Os (g) + H20 (1) ► 2 HNO (1) AH = -76.6 kJ N2(g) + 3 O2 (g) + H2(g) → 2 HNO, (1) AH = -348.2 kJ a. Calculate the AHⓇ for the reaction: 2 N2 (g) + 5 O2(g) → 2 N2Os (g)