Calculate Δ U of reaction for the hydrogenation of acetylene at constant volume and at 77ºC. Given that Δ H f (H 2 O) = –67.8 kcal mole; Δ H comb (C 2 H 2 ) = –310.1 kcal/mole, Δ H comb (C 2 H 4 ) = –337.2 kcal/Mole
Text Solution
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Required equation is :
C 2 H 2(g) + H 2(g) ⎯→ C 2 H 4(g) Δ H – = Δ U – = ?
H 2(g) + 
O 2(g) ⎯→ H 2 O( λ ) Δ H – = –67.8 ...
C 2 H 2(g) + 
O 2(g) ⎯→ 2CO 2 + H 2 O( λ ) Δ H – = –310.1 ...
C 2 H 4 + 3O 2(g) ⎯→ 2CO 2(g) + 2H 2 O( λ ) Δ H – = –337.2 ...
+ –
⇒ C 2 H 2(g) + H 2(g) ⎯→ C 2 H 4(g)
Δ H r×n = –67.8 – 310.1 + 337.2 = –40.7 KCal
Δ H = Δ U + Δ n (g) RT
–40.7 = Δ U + (–1) × 2 × 10 –3 × 350
Δ U = –40.7 + .7
Δ U = –40 KCal/mole
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