Calculate ΔHrxno (kJ/mole) for the reaction
C2H4(g) + 6 F2(g) ----- > 2 CF4(g) + 4 HF(g) ΔHrxno (kJ/mole) = ?
Using only the following data ∆Hºrxn(kJ/mole)
H2(g) + F2(g) ------ > 2 HF(g) -537.0
C(s) + 2 F2(g) ----- > CF4(g) - 680.0
2 C(s) + 2 H2(g) ----- > C2H4(g) + 52.3
Lets number the reaction as 0, 1, 2, 3 from top to bottom
required reaction should be written in terms of other reaction
This is Hess Law
required reaction can be written as:
reaction 0 = +2 * (reaction 1) +2 * (reaction 2) -1 * (reaction 3)
So, ΔHo rxn for required reaction will be:
ΔHo rxn = +2 * ΔHo rxn(reaction 1) +2 * ΔHo rxn(reaction 2) -1 * ΔHo rxn(reaction 3)
= +2 * (-537.0) +2 * (-680.0) -1 * (52.3)
= -2486.3 KJ
Answer: -2486.3 KJ
4. (14 pts) Calculate AHr (kJ/mole) for the reaction CaHlg)+ 6 F2(g) >2 CFg) + 4 HF(g) AH (kJ/mole) ? Using only the following data AH (kJ/mole) H2(g)+ F2(g) C(s) + 2 F2(g) 2 C(s)+ 2 H2(g) 2 HF(g) -537.0 -680,0 CF4(g) C2H4(g) + 52.3
Calculate the standard enthalpy of formation of gaseous carbon tetrafluoride (CF4) using the following thermochemical information: 2 HF(g) H2(g) + F2(g) H = +537 kJ 2 CF4(g) + 4 HF(g) C2H4(g) + 6 F2(g) H = +2486.3 kJ C2H4(g) 2 C(s) + 2 H2(g) H = -52.3 kJ Calculate the standard enthalpy of formation of gaseous carbon tetrafluoride (CF4) using the following thermochemical information: 2 HF(g) H2(g) + F2(g) H = +537 kJ 2 CF4(g) + 4 HF(g) C2H4(g) +...
3.(15pts) In each case, find the missing AH. (kJ/mole) (a) If 2 Al(s) + Fe2O3(s) ----> Al2O3(s) + 2 Fe(s) AH,xº = -851.5kJ/mol then 1/8 Al2O3(s) + 1/4 Fe(s) --> 1/4 Al(s) + 1/8 Fe2O3(s) AHX = ? (b) If CaO(8) + 3 C(s) --> CaC2(s) + CO(g) AH,” = 464,8 kJ/mol then 6 CaO(s) + 18 C(s) ---> 6 CaCz(s) +6 CO(g) AHX = ? (c) If N2(g) + 2 O2(g) ---> N204(g) and 2 NO2(g) ----> N2(g) +...
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
Calculate the standard enthalpy of formation of gaseous hydrogen fluoride (HF) using the following thermochemical information: C2H4(g) + 6 F2(g) 2 CF4(g) + 4 HF(g) H = -2486.3 kJ CF4(g) C(s) + 2 F2(g) H = +680 kJ C2H4(g) 2 C(s) + 2 H2(g) H = -52.3 kJ H = ??? kJ
Calculate the standard enthalpy of formation of gaseous hydrogen fluoride (HF) using the following thermochemical information: C2H4(g) + 6 F2(g) 2 CF4(g) + 4 HF(g) H = -2486.3 kJ CF4(g) C(s) + 2 F2(g) H = +680 kJ C2H4(g) 2 C(s) + 2 H2(g) H = -52.3 kJ H = ___?kJ The answer is not -589.3
Calculate the standard enthalpy of formation of gaseous hydrogen fluoride (HF) using the following thermochemical information: C2H4(g) + 6 F2(g) 2 CF4(g) + 4 HF(g) (delta)H = -2486.3 kJ CF4(g) C(s) + 2 F2(g) (delta)H = +680 kJ 2 C(s) + 2 H2(g) C2H4(g) (delta)H = +52.3 kJ (delta)H =__________ 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)
Practice with Hess's Law and Standard Heats of Formation 1. (Example) The reaction C2H4 (g) + 6 F2 (g) → 2 CF4(g) + 4 HF (g) can be written as the sum of: C2H4 (9) ► 2 C(s) + 2 H2 (g) AH = -52.3 kJ/mol 2 C(s) + 4 F2 (9) ► 2 CF4(9) AH = -1360 kJ/mol 2 H2(g) + 2 F2 (g) → 4 HF (a) AH = -1074 kJ/mol C2H4(g) + 6 F2(g) → 2 CF4(g)...
What is the enthalpy of reaction, dH(rxn), for the following reaction? C2H4 (g). +. 6 F2 (g). --->. 2 CF4 (g). +.4 HF (g) dH(f) +52.3 -680 -268.5. kJ/mole dH(rxn) - +2486 kJ dH(rxn) = -2486 kJ o dH(rxn) = -2434 kJ • dH(rxn) = -1000.8 kJ