ΔH° = –484 kJ 0.25 mol O2 0.50 mol H2 –5.6 L PV work ΔE

ΔH° = –484 kJ 0.25 mol O2 0.50 mol H2 –5.6 L PV work ΔE

 

 

The reaction between hydrogen and oxygen

to yield water vapour has ΔH° = –484 kJ.

2H2(g) + O2(g) → 2H2O(g) ... ΔH° = –484 kJ

How much PV work is done and what is the value of ΔE (in kJ)

for the reaction of 0.50 mol of H2 and 0.25 mol of O2

at atmospheric pressure if the volume change is –5.6 L?

 

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w = –PΔV

( 참고 https://ywpop.tistory.com/5281 )

 

= –(1 atm) (–5.6 L)

= 5.6 atm•L

 

 

 

1 atm•L = 101.325 J 이므로,

( 참고 https://ywpop.tistory.com/6646 )

 

5.6 atm•L × (101.325 J / 1 atm•L)

= 567 J

 

 

 

2H2(g) + O2(g) → 2H2O(g)

 

H2 : O2 = 2 : 1 계수비(= 몰수비) 이므로,

0.50 mol H2와 0.25 mol O2는 전부 반응.

 

 

 

2 mol H2 반응 시, ΔH° = –484 kJ 이므로,

0.50 mol H2 반응했으니,

0.50 mol × (–484 kJ / 2 mol)

= –121 kJ

 

 

 

ΔE = q + w

( 참고 https://ywpop.tistory.com/5134 )

 

= (–121000) + (567)

= –120433 J

= –120.4 kJ

 

 

 

답: –120 kJ