Lets number the reaction as 1, 2, 3, 4 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 4 = +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
6.65 From the enthalpies of reaction H2(8) + F2(8) 2 HF(8) AH = -537 kJ C(s)...
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...
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)
2. Calculate AH for the reaction of ethylene with F2. CHA(g) + 6 F2(g) → 2 CF.(g) + 4 HF(g) using the following enthalpies of reaction: 2 H2(g) + 2 F2(g) → 4 HF(g) AH, = -1074 kJ C(s) + 2 F2(g) → CF4(g) AH2 = -680 kJ C(s) + H2(g) → CH.(g) AH, = 26.2 kJ
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
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: 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 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) +...
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