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

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Solution

The reaction can be written as
$C_{2} H_{4} + H_{2} ⟶ C_{2} H_{6}$
The ratio of reactants is $1 : 1$ and product is also forming in ratio of $1$
$∴ 50 m l$ of ethylene on reaction with $50 m l$ of $H_{2}$ will produce $50 m l$ of ethane.
$\Rightarrow$ Change in volume of ethane is
$50 m l (e t h a n e) - 100 m l (C_{2} H_{4} + H_{2}) = 50 m l = 0.05 l t$
$P △ V = 1.5 \times 0.05 = 0.075 l$ $a t m = 7.5$ joules
$△ H = △ U + P △ V$
$△ U = △ H - P △ V$
$= (- 0.31) - 1.5 (0.005)$
$= - 0.31 - 0.075$
$= - 0.385$

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The enthalpy change for the reaction of 50ml of ethylene with 50.0ml of H2 at 1.5 atm pressure is ΔH=−0.31 KJ. What is the ΔU?

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