The Enthalpy change for the reaction of $59 m l$ of ethylene with $50.0 m l$ of $H_{2}$ at $1.5$ bar pressure is $\Delta H=-0.31\ KJ$. What is the change in internal energy?

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Solution

$C_{2} H_{4} + H_{2} ⟶ C_{2} H_{6}$
As from the above reaction,
$1$ mole of ethylene requires $1$ mole of hydrogen to react.
Given volume of ethylene $= 59 m L$
Given volume of hydrogen $= 50 m L$
$∴$ Only $50 m L$ of ethylene will react.
Pressure $(P) = 1.5 b a r = 1.48 a t m$
$∴ W = P Δ V = 1.48 \times 0.05 = 0.074 L a t m = (0.074 \times 101.3) J = 7.496 J$
From first law of thermodynamics-
$Δ U = Δ H + W$
Given that $Δ H$ for the reaction is $- 0.31 K J = - 310 J$ .
$∴ Δ U = - 310 + 7.496 = 302.504 J = 0.302 K J$
Hence the correct answer is $0.302 K J$ .

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The Enthalpy change for the reaction of 59ml of ethylene with 50.0ml of H2 at 1.5 bar pressure is $\Delta H=-0.31\ KJ$. What is the change in internal energy?

The Enthalpy change for the reaction of $59 m l$ of ethylene with $50.0 m l$ of $H_{2}$ at $1.5$ bar pressure is $\Delta H=-0.31\ KJ$. What is the change in internal energy?