Calculate AH for the reaction CyHa(s) + 6F2(8) 2CF4(8) + 4HF(G) given the following information: As...
From the enthalpies of reactio:
H2(
9.) From the enthalpies of reaction Hale) + F(b) → C(s) + 2F2(g) → 2C(s) + 2H2(g) → Calculate AH for the reaction of ethylene with F2: CzHa() + 6F2(g) 2HF(8) CF(b) CzH4(g) AH =-537 kJ AH = -680 kJ AH = +52.3 kJ → 2CF4(g) + 4HF(E)
can you explain why the answer is correct?
QUESTION 11 Given the information below calculate AHrxn for: C2Halg)+6F2g)> 2CF4 + 4HF(g) => 2HF(g ) AH°= -537 kJ ->CF4(g ) A °H-680 kJ H2g)+F2(g) C(s)+2F29) 2C(s)+2H2(g)=> C2H4(g) AH 52.3 kJ A 1165 kJ B..1269 C.-2382 kJ D.-2486 kJ E. 1642 kJ
6.65 From the enthalpies of reaction H2(8) + F2(8) 2 HF(8) AH = -537 kJ C(s) + 2 F2(8) CF4(8) AH = -680 kJ 2 C(s) + 2 H2(8) —— C2H4(8) AH = +52.3 kJ calculate AH for the reaction of ethylene with F2: C2H4(8) + 6 F2(8) — 2 CF4(8) + 4 HF(8)
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
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)
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
Calculate the standard enthalpy of formation of gaseous carbon
tetrafluoride (CF4) using the following thermochemical
information:
Calculate the standard enthalpy of formation of gaseous carbon tetrafluoride (CF4) using the following thermochemical information: H2(g)F2(g) 2 HF(g) 2 CF4(g)4 HF(g) C2H4(g) +6 F2(g) C2H4(g)2 C(s) 2 H2(g) AH -537 kJ AH 2486.3 kJ AH -52.3 kJ ΔΗ - kJ
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
Question 8 of 20 Using the equations H2(g) + F2 (g) → 2 HF (g) AH° = -79.2 kJ/mol C(s) + 2 F2 (g) → CF. (g) AH° = 141.3 kJ/mol 2 C(s) + 2 H2 (9) C2H4 (9) AH = -97.6 kJ/mol Determine the enthalpy for the reaction C2H4 (g) + 6 F2 (g) → 2 CF4 (9) + 4 HF (g). kJ/mol
5.65 From the enthalpies of reaction H 2 (g)+r 2 (g)
longrightarrow2 HF(s) Delta*H = - 537kJ C(s)+2 F 2 (s)
longrightarrow CF 4 (g) Delta*H = - 680kJ 2C(s)+2 H 2 (g) C 2 H 4
(g) Delta*H = + 52.3kJ calculate Delta*H for the reaction of
ethylene with F 2 : C 2 H 4 (g)+6 F 2 (g) 2 CF 4 (g)+4 HF(g)
5.65 From the enthalpies of reaction H2(8) + (8) 2 HF(8) AH = -537...