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The enthaly change for the reaction...

The enthaly change for the reaction
`C_(2) H_(4((g))) + H_(2_((g))) to C_(2) H_(6_((g)))`
is - 620 J when 100 mL of ethylene and 100 mL of `H_(2)` react at 1 atm pressure .Calculate the pressure volume work and `DeltaU`.

Text Solution

Verified by Experts

The correct Answer is:
`W= 10.13 J ; Delta U =- 609 .87 J`
Given : `DeltaH = - 620 J`
Volume of `C_(2) H_(4_((g))) ` taken = 10 mL
Volume of `H_(2_((g))) ` taken = 100 mL
`P_(ex) = 1 atm`
`W= ? Delta U = ?`
`C_(2) H_(4_((g))) + H_(2_((g))) to C_(2) H_(6_((g)))`
`"At start " " "100 mL " "100 mL " "0`
`"After reaction " " "0" "0 " " 100 mL`
(By Gay- Lussac's law)
Initial volume `=V_(1) = 100 + 100 = 200 mL`
Final volume `= V_(2) = 100 mL`
`:. Delta V = V_(2) = V_(1) = 100 - 200 =- 100 mL =- 0.1 L`
`W=- P _(ex) (V_(2)-V_(1))`
`= - P_(ex) Delta V`
`=- 1 xx (-0.1)`
`=+ 0.1 L "atm"`
`=0.1 xx 101 .3 J ('1 L "atm" = 101.3 J)= 10.13 J`
This is a work of compression due to decrease in volume
`Delta H= Delta U + P Delta V`
`:. Delta U = Delta H - P Delta V =- 620 -(- 10. 13) =- 609 .87 kJ`
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