The heat of combustion of ethylene at $18^{o} C$ and at constant volume is $- 335.8 k c a l$ when water is obtained in liquid state. Calculate the heat combustion at constant pressure and at $18^{o} C$ .

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Solution

The chemical equation for the combustion of $C_{2} H_{4}$ is
$C_{2} H_{4} (g) 1 m o l e + 3 O_{2} (g) 1 m o l e = 2 {C O}_{2} (g) 2 m o l e s + 2 H_{2} O (l); Δ U = - 335.8 k c a l$
No. of moles of reactants $= (1 + 3) = 4$
No. of moles of products $= 2$
So, $Δ n = (2 - 4) = - 2$
Given, $Δ U = - 335.8 k c a l, Δ n = - 2, R = 2 \times 10^{- 3} k c a l$
and $T = (18 + 273) = 291 K$
Applying $Δ H = Δ U + Δ n R T = - 335 + (- 2) (2 \times 10^{- 3}) (192) = - 336.964 k c a l$

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The heat of combustion of ethylene at 18oC and at constant volume is −335.8kcal when water is obtained in liquid state. Calculate the heat combustion at constant pressure and at 18oC.

The heat of combustion of ethylene at $18^{o} C$ and at constant volume is $- 335.8 k c a l$ when water is obtained in liquid state. Calculate the heat combustion at constant pressure and at $18^{o} C$ .