{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Split_00005

# Split_00005 - 8.15(cont'd iii Q= 250 mol 2.54 20.91[kJ mol...

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

8- 5 8.15 (cont’d) iii) ± . . . Q = = − 250 2 54 20 91 1276 mol 3600 s [kJ / mol] kW b g b. Method (i) is most accurate since it takes into account the dependence of enthalpy on pressure and (ii) and (iii) do not. c. The enthalpy change for steam going from 10 bar to 1 atm at 600 o C. 8.16 Assume ideal gas behavior, so that pressure changes do not affect Δ ² H . ± . . n R R = = 200 492 12 1 0 6125 ft h 537 atm 1 atm lb - mol 359 ft (STP) lb - mole / h 3 o o 3 ± ± ² ( . ( ) Q n H = = = Δ 0 6125 2993 0 1833 lb - mole h ) [Btu / lb - mole] Btu / h b g 8.17 a. 50 kg 1.14 kJ 50 C kg C kJ ° ⋅° = 10 2280 b g b. ( ) ( ) ( ) ( ) ( ) ( ) 2 3 Na CO Na C O 2 3 2 0.026 0.0075 3 0.017 0.1105 kJ mol C p p p p C C C C + + = + + = ⋅° 50,000 g 0.1105 kJ 1 mol 50 C mol C 105.99 g 2085 kJ % error error ° ⋅° = = × = − 10 2085 2280 2280 100% 8 6% b g . 8.18 C p d i b g b g b g C H O(l) o 6 14 kJ / (mol C) = + + = 6 0 012 14 0 018 1 0 025 0 349 . . . . (Kopp’s Rule) C p d i CH COCH (l) 3 3 T kJ (mol C = + × ⋅° 01230 18 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}