Calculate the enthalpy of the following reaction. H2C = CH2(g) + H2(g) →CH3 − CH3(g)
The bond energies of C - H, C - C, C = C & H - H are 99, 83, 147 & 104 Kcal respectively.
30 K cal
-30 K cal
20 K cal
-20 K cal
Bond energy for C-H bond & C-Cl bond are 415KJ and 326KJ respectively.
0.5 K cal
-0.5 K cal
-188.5 K cal
188.5 K cal
Bond energy of N ≡ N, N = N, O = O and N = O bonds are 946, 418, 498 & 607 kJ mol−1 respectively.
- Conservation of mass
- First law of thermodynamics
- None of the above
- All of these
Given the bond energies N ≡ N, H − H and N − H bonds are 945, 436 and 391 kJ mole−1 respectively, the enthalpy of the following reaction N2(g) + 3H2(g) → 2NH3(g) is
If at 298 K the bond energies of C - H, C - C, C=C and H - H bonds are respectively 414, 347, 615 and 435 kJ mol−1, the value of enthalpy change for the reaction
The molar entropies of HI(g), H(g) and I(g) at 298K are 206.5, 114.6, and 180.7 J mole−1 K−1 respectively. Using the Δ G0 given below, Calculate the bond energy of HI.
H(g) → H(g) + I(g); Δ G0 = 271.8 kJ
Ethyl chloride (C2H5Cl), is prepared by reaction of ethylene with hydrogen chloride :
C2H4(g)+HCl(g)→C2H5Cl(g) ΔH=−72.3kJ. What is the value of ΔE (in KJ), if 70g of ethylene and 73g of HCl are allowed to react at 300K?