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Transcript
214
c)
ΔHº = –393.5 kJ
C(graphite) + O 2 (g) ⎯⎯
→ CO 2 (g) ,
Given :
i)
C 2 H 6 (g) +7/2O 2 (g) ⎯⎯
→ 2CO 2 (g)+3H 2O (l ) ,
ΔHº = –1560 kJ
ii)
H 2 (g) + 1/2O 2 (g) ⎯⎯
→ H 2 O(l ) ,
ΔHº = –285.8 kJ
ΔHº = –393.5 kJ
iii) C(graphite) + O 2 (g) ⎯⎯
→ CO 2 (g) ,
To find :
ΔHº(reaction) = ?
Solution :
→ C 2 H 6 (g) ; ΔHº = ?
Required equation : 2C (graphite) + 3H 2 (g) ⎯⎯
Treatment : Reverse eq. (i) + 3 × eq. (ii) + 2 × eq. (iii)
7
2CO 2 (g) + 3H 2O(l ) ⎯⎯
→ C 2 H 6 + O 2 (g) , ΔHº = + 1560 kJ
2
3
3H 2 (g) + O 2 (g) ⎯⎯
→ 3H 2 O(l ) ,
ΔHº = –857.4 kJ
2
2C(graphite) + 2O 2 (g) ⎯⎯
→ 2CO 2 (g) ,
ΔHº = –787.0 kJ
2C(graphite) + 3H 2 (g) ⎯⎯
→ C 2 H 6 (g) , ΔHº = +1560 kJ – 857.4 kJ – 787 kJ
Δ Hº = –84.4 kJ.
*21.
Given the following equations calculate the standard enthalpy of the reaction,
2Fe(s) + 3/ 2 O 2 (g) ⎯⎯
→ Fe 2O 3 (s) ,
ΔHº = ?
a)
2Al(s) + Fe 2O 3 (s) ⎯⎯
→ 2Fe(s) + Al 2 O 3 (s) , ΔHº = –847.6 kJ
b)
2Al(s) + 3/2 O 2 (g) ⎯⎯
→ Al 2 O 3 (s) ,
ΔHº = –1670 kJ
Given :
i)
2Al(s) + Fe 2O 3 (s) ⎯⎯
→ 2Fe(s) + Al 2 O 3 (s) , ΔHº = –847.6 kJ
ii) 2Al(s) + 3/2 O 2 (g) ⎯⎯
→ Al 2 O 3 (s) ,
To find :
ΔHº(reaction) = ?
Solution :
Required equation : 2Fe(s) +
ΔHº = –1670 kJ
3
O (g) ⎯⎯
→ Fe 2O 3 (s)
2 2
Treatment : Reverse eq. (i) + eq. (ii)
Unique Solutions ®
S.Y.J.C. Science - Chemistry - Part I