Bond Energies (kJ/mol)BondBond EnergyBondBond EnergyBondBond EnergyH–H436C–S260F–Cl255H–C415C–Cl330F–Br235H–N390C–Br275Si–Si230H–O464C–I240Si–P215H–F569N–N160Si–S225H–Si395N=NN=N418Si–Cl359H–P320N≡NN≡N946Si–Br290H–S340N–O200Si–I215H–Cl432N–F270P–P215H–Br370N–P210P–S230H–I295N–Cl200P–Cl330C–C345N–Br245P–Br270C=CC=C611O–O140P–I215C≡CC≡C837O=OO=O498S–S215C–N290O–F160S–Cl250C=NC=N615O–Si370S–Br215C≡NC≡N891O–P350Cl–Cl243C–O350O–Cl205Cl–Br220C=OC=O741O–I200Cl–I210C≡OC≡O1080F–F160Br–Br190C–F439F–Si540Br–I180C–Si360F–P489I–I150C–P 265F–S285Molecular ekacagine combines with ethylene to form diekaethylene (C2H2Ec2) according to the reaction below: Examine the bond energies in Table 9.3 of your textbook. Shuri’s undergrads measured the new bond enthalpies: C-Ec bond energy is 220 kJ/mol and the Ec-Ec bond energy is 120 kJ/mol. Calculate the enthalpy change when 1 mol of ethylene is converted to diekaethylene.Calculate the enthalpy change when 9.00 x 103 kg of ethylene is converted to diekaethylene. HHH.H--C-C.Ec-Ec +H.ÉcÉc

In a chemical reaction, some bonds in reactants are broken (which requires energy I.e. + sign) and some bonds are formed to give products (which release energy i.e. - sign) So enthalpy of a reaction = sum of bond energy of reactants - sum of bond energy of products ΔHrxn = ∑ Bond energy of reactants - ∑ Bond energy of products.(1) In reactants, there is 1 C=C and 4 C-H bonds in ethylene, also there is an Ec-Ec bond to be broken. In product Ethane, there is 1 C-C and 4 C-H bonds and 2 C-Ec bonds. ∆H = (1×611)+(4×415)+120-345+(4×415)+(2×220)∆H =611+1660+120 -345 + 1660+440=2391-2445=-54 kJ/mol(b) mass of ethylene = 9000 Kg = 9000 × 103 g. The no of moles of ethylene = n =9000×103 g28 g/mol=321428.571 mol of ethylene produces 321428.571 mol of diekaethyleneEnthalpy change =321428.571 mol ×-54 kJ/mol=- 17357142.8 kJ

Answered: Molecular ekacagine combines with…

Chemistry

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ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Bond Energies (kJ/mol)
Bond	Bond Energy	Bond	Bond Energy	Bond	Bond Energy
H–H	436	C–S	260	F–Cl	255
H–C	415	C–Cl	330	F–Br	235
H–N	390	C–Br	275	Si–Si	230
H–O	464	C–I	240	Si–P	215
H–F	569	N–N	160	Si–S	225
H–Si	395	N=NN=N	418	Si–Cl	359
H–P	320	N≡NN≡N	946	Si–Br	290
H–S	340	N–O	200	Si–I	215
H–Cl	432	N–F	270	P–P	215
H–Br	370	N–P	210	P–S	230
H–I	295	N–Cl	200	P–Cl	330
C–C	345	N–Br	245	P–Br	270
C=CC=C	611	O–O	140	P–I	215
C≡CC≡C	837	O=OO=O	498	S–S	215
C–N	290	O–F	160	S–Cl	250
C=NC=N	615	O–Si	370	S–Br	215
C≡NC≡N	891	O–P	350	Cl–Cl	243
C–O	350	O–Cl	205	Cl–Br	220
C=OC=O	741	O–I	200	Cl–I	210
C≡OC≡O	1080	F–F	160	Br–Br	190
C–F	439	F–Si	540	Br–I	180
C–Si	360	F–P	489	I–I	150
C–P	265	F–S	285

Molecular ekacagine combines with ethylene to form diekaethylene (C₂H₂Ec₂) according to the reaction below:

Examine the bond energies in Table 9.3 of your textbook. Shuri’s undergrads measured the new bond enthalpies: C-Ec bond energy is 220 kJ/mol and the Ec-Ec bond energy is 120 kJ/mol. Calculate the enthalpy change when 1 mol of ethylene is converted to diekaethylene.
Calculate the enthalpy change when 9.00 x 10³ kg of ethylene is converted to diekaethylene.

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