Split_00083 - 9.65 (cont'd) substance C5 H 12 O2 CO 2 H2O...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
9- 83 substance mol kJ mol mol kJ mol CH O CO HO in in out out 51 2 2 n H n H HH H H ±± . . ± . ± . ± . ± 100 0 10 40 2 40 500 600 21 2 23 4 −− ± () , ± HC d T i d T ip i T p T == =+ z z C for H O(v) v o (v ) 2 2 25 25 25 Δ ej (. 1 75 148 = B O o Table B.8 2 C) = kJ / mol Substituting ( ) from Table B.2 : kJ mol kJ mol C HT T T T T T T pi ad ad ad ad ad ad ad ad ± . . . . ) ± . . . . ) 2 5 2 8 3 12 4 3 5 2 8 3 12 4 0 0291 0579 10 0 2025 10 0 3278 10 0 7311 0 03611 21165 10 0 9623 10 1866 10 0 9158 × × + × × × + × −−− ± .( . . . . . ) T TT ad ad ad ad 4 5 2 8 3 12 4 44 01 003346 03440 10 02535 10 08983 10 0838 + × + × × kJ mol ⇒= + + × + × × ± . . . . T ad ad ad ad 4 5 2 8 3 12 4 4317 003346 03440 10 02535 10 08983 10 ) Energy balance : Δ H = 0 n Hn H n H ii c o CH l ) out in 2 2 ± ( +−= ( . ) ( . ) ± (. ) ± ± (.) ( ± ) 1 35095 240 1040 0 234 1 mol C H kJ / mol 2 −+ + + = HHH H Substitute for through ad ad ad ad ad ad ad ad ad ad ad o kJ / mol = 0 Check : Solving for using E- Z Solve C ² . . . ) . . . . . . . . . T T T fT
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.
Ask a homework question - tutors are online