The enthalpy change for the reaction of 50 ml of ethylene with 50.0 ml of $H_{2}$ at 1.5 atm pressure $△ H = - 0.31 k J$ . What is $△ U$ ?

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Solution

$C_{2} H_{4} Ethylene + H_{2} ⟶ C_{2} H_{6}$
From the reaction,
$1$ mole of ethylene reacts with $1$ mole of $H_{2}$ to produce $1$ mole of product.
Thus $50 m L$ of ethylene reacts with $50 m L$ of $H_{2}$ to produce $50 m L$ of product.
Therefore,
Initial volume $(V_{1}) = 50 + 50 = 100 m L$
Final volume after forming $50 m L$ of product $(V_{2}) = 50 m L$
$Δ V = 50 - 100 = - 50 m L = - 0.05 L$
Given pressure $(P) = 1.5 a t m$
therefore,
$P Δ V = - 1.5 \times (- 0.05) = - 0.075 L - a t m = - 7.6 J$
$Δ H = - 0.31 K J = 310 J (Given)$
$∴ Δ U = Δ H - P Δ V = - 310 - (- 7.6) = - 302.4 J = - 0.3024 K J$
Hence $Δ U$ for the reaction will be $- 0.3024 K J$ .

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The enthalpy change for the reaction of 50 ml of ethylene with 50.0 ml of H2 at 1.5 atm pressure △H=−0.31kJ. What is △U?

The enthalpy change for the reaction of 50 ml of ethylene with 50.0 ml of $H_{2}$ at 1.5 atm pressure $△ H = - 0.31 k J$ . What is $△ U$ ?