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

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(...
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)
Calculate the standard enthalpy of reaction: 2F2 (?) + 2H2O(?) → 4HF(?) + O2 (?). ∆?°? (F2, ?) = 0 kJ mol ∆?°? (O2, ?) = 0 kJ/mol ∆?°? (HF,?) = −268.61 kJ/mol ∆?°? (H2O, ?) = −285.8 kJ/mol
A chemist knows that the
kJ for the reaction 2H2(g) + O2
(g) ---> 2H2O (g) ,and that
kJ
for the reaction H2 (g) + F2 (g) ---> 2HF
(g).
With this information he calculated the
for the reaction 2H2O (g) + 2F2
(g) ---> 4HF(g) + O2 (g) and
predicted whether
was positive or negative. How?
A Ho- 485
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
Hess's Law Practice Find AH° for the following equation: SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(g) Using the following equations: Si(s) + O2(g) → SiO2(s) Si(s) + 2F2(g) → SiF4(g) H2(g) + F2(g) → 2HF(g) H2(g) + 4202(g) → H2O(g) AH = -910.9 kJ/mol rxn AH = -1651 kJ/mol rxn AH = -542 kJ/mol rxn AH = -241.8 kJ/mol rxn
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-'
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 of AHixn for 2F2(g) + 2 H2O(l) — 4 HF(g) + O2(8)
Use Hess’s law to calculate ∆H° for the reaction:C(s) + 2H2(g) + ½O2(g) → CH3OH(l) ∆H°∘= ?using only the following data:H2(g) + ½O2(g) → H2O(l) ∆H°= -285.8 kJC(s) + O2(g) → CO2(g) ∆H°= -393.5 kJ2CH3OH(g) + 3O2(g) → 2CO2(g) + 4H2O(l) ∆H°= -1452.8 kJ
Hess’s Law (a) Calculate the ∆H for the reaction: CCl4(g) → C(s, graphite) + 2Cl2(g) using the following chemical equations and their respective enthalpy changes: C(s, graphite) + 2F2(g) → CF4(g) ∆H = -679.9 kJ CF4(g) + 2Cl2(g) → CCl4(g) + 2F2(g) ∆H = 573.2 kJ (b) Calculate the ∆H for the reaction: C(s, graphite) + 2H2O(g) → CH4(g) + O2(g) using the following chemical equations and their respective enthalpy changes: C(s, graphite) + O2(g) → CO2(g) ∆H = -394...
Thermochemical equations 5. Given 2NO → N2 + O2 ∆H= -180.7 determine the enthalpy of the reverse reaction? Is the reverse reaction endothermic or exothermic? 6. Given H2 + F2 → 2HF ∆H= -537 kJ a) How much heat is required to react 9.5 g F2 with H2? b) What mass of H2 is needed to react with F2 with -294 kJ of energy? Hess Law State Hess’s Law 8. Use the standard reaction enthalpies given below to determine ΔH°...