1. How many moles of O, are needed to burn 58.7g of propane (C2H3) according to the
1 answer:
You might be interested in
Answer and Explanation:
ΔH = (Bond energies of the products) - (Bond Energies of the reactants)
a) C2H6 + H2 ----> 2CH4 ΔH = -65.07 KJ/mol
Bond energy of product = 2×Bond energy of CH4 =2 × (4 (C-H)) =2 × 4 × 415 = 3320 KJ/mol
Bond energy of reactants = Bond energy of C2H6 + Bond energy of H2
Bond energy of C2H6 = (C-C) + 6(C-H) = 347 + 6(C-H)
Bond energy of H2 = (H-H) = 432 KJ/mol
-65.07 = 3320 - (Bond energy of C2H6 + 432)
Bond energy of C2H6 = 3320-432+65.07 = 2953.07 KJ/mol
Bond energy of C2H6 = (C-C) + 6(C-H) = 347 + 6(C-H) = 2953.07
(C-H) = (2953.07 -347)/6 = 434.345 KJ/mol
b) C2H4 + 2H2 ------> CH4 ΔH = -202.21 KJ/mol
Bond energy of product = 2×Bond energy of CH4 =2 × (4 (C-H)) =2 × 4 × 415 = 3320 KJ/mol
Bond energy of reactants = Bond energy of C2H4 + (2 × Bond energy of H2)
Bond energy of C2H4 = (C=C) + 4(C-H) = 614 + 4(C-H)
(2 × Bond energy of H2) = 2 × 432 = 864 KJ/mol
-202.21 = 3320 - (Bond energy of C2H4 + 864)
Bond energy of C2H4 = 3320+202.21-864 = 2658.1 KJ/mol
Bond energy of C2H4 = (C=C) + 4(C-H) = 614 + 4(C-H) = 2658.1
(C-H) = (2658.1 - 614)/4 = 511.05 KJ/mol
c) C2H2 + 3H2 ------> 2CH4 ΔH = -376.74 KJ/mol
Bond energy of product = 2×Bond energy of CH4 =2 × (4 (C-H)) =2 × 4 × 415 = 3320 KJ/mol
Bond energy of reactants = Bond energy of C2H2 + (3 × Bond energy of H2)
Bond energy of C2H = (triple bond Carbon to Carbon) + 2(C-H) = 839 + 2(C-H)
(3 × Bond energy of H2) = 3 × 432 = 1296 KJ/mol
-376.74 = 3320 - (Bond energy of C2H2 + 1296)
Bond energy of C2H2 = 3320+376.74-1296 = 2400.74 KJ/mol
Bond energy of C2H2 = (triple bond Carbon to Carbon) + 2(C-H) = 839 + 2(C-H) = 2400.74
(C-H) = (2400.74 - 839)/2 = 780.87 KJ/mol
QED!!!
