Answer to Question #152366 in Physical Chemistry for naphtali

Question

To estimate the standard enthalpy of formation of acetone vapour. Given the following information :
B (H - H) : 436 kJ mol⁻¹
B (O = O) : 497 kJ mol⁻¹
B (C - H) : 412 kJ mol⁻¹
B (C - C) : 348 kJ mol⁻¹
B (C = O) : 745 kJ mol⁻¹, ∆_atom H (graphite) = 717 kJ mol⁻¹.

Answer 1

O

                                      ∣∣

3C(g)+3H2​(g)+1/2​O2​(g)→CH3​−C−CH3​ 

Heat of formation of Acetone = [Bond energy of formation of bond]

+ [Bond energy of dissociation of bond]

=[6(C−H) +2(C−C) +1(C=0)]

+[3{C(s)→C(g)} +3(H−H) +1​/2(0−0)]

=− [6×412+2×348+1×745]

+ [3×717+3×436+1/2​×497]

=− [2472+696+745]

+ [2151+1308+248.5]

=−3913kJ+3707.5kJ

=−205.5kJ/mol

∴ Heat of formation of acetone is −205.5kJ/mol