C2H6g+72O2g→2CO2g+3H2OlΔH=−1560KJ;C2H2g+52O2g→2CO2g+H2Ol ΔH=−1310KJ;H2g+12O2g→H2Ol ΔH=−290KJ;Enthalpy of hydrogenation of ethyne (C2H2) is

Required equation ; C2H2g+2H2g→C2H6gΔH=Eq(2)+2Eq(3)−Eq(1)ΔH=−1310+2(−290)−(−1560)ΔH=−330KJ

C2H6g+72O2g→2CO2g+3H2OlΔH=−1560KJ;C2H2g+52O2g→2CO2g+H2Ol ΔH=−1310KJ;H2g+12O2g→H2Ol ΔH=−290KJ;Enthalpy of hydrogenation of ethyne (C2H2) is

Required equation ; C2H2g+2H2g→C2H6gΔH=Eq(2)+2Eq(3)−Eq(1)ΔH=−1310+2(−290)−(−1560)ΔH=−330KJ

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$C_{2} H_{6} (g) + \frac{7}{2} O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (l) Δ H = - 1560 KJ;$
$C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) \to 2 C O_{2} (g) + H_{2} O (l) Δ H = - 1310 KJ;$
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) Δ H = - 290 KJ;$
Enthalpy of hydrogenation of ethyne (C₂H₂) is

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-40 KJ

-620 KJ

-330 KJ

-410 KJ

answer is C.

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Detailed Solution

Required equation ;
$C_{2} H_{2} (g) + 2 H_{2} (g) \to C_{2} H_{6} (g)$
$\begin{array}{l} Δ H = E q (2) + 2 E q (3) - E q (1) \\ Δ H = - 1310 + 2 (- 290) - (- 1560) \\ Δ H = - 330 K J \end{array}$

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C2H6g+72O2g→2CO2g+3H2OlΔH=−1560KJ;C2H2g+52O2g→2CO2g+H2Ol ΔH=−1310KJ;H2g+12O2g→H2Ol ΔH=−290KJ;Enthalpy of hydrogenation of ethyne (C2H2) is

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