Ethyl chloride $(C_{2} H_{5} Cl)$ is prepared by reaction of ethylene With hydrogen chloride:
$C_{2} H_{4} (g) + HCl (g) ⟶ C_{2} H_{5} Cl (g) Δ H = - 72.3 kJ / mol .$
What is the value of $Δ E$ (in KJ) if 98 g of ethylene and 109.5 g of HCl are allowed to react at 300 K ?

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-64.81

-190.71

-209.41

-224.38

answer is C.

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Detailed Solution

We know that,

$\begin{array}{l} Δ H = Δ E + Δ n_{g} R T \\ - 72.3 = Δ E - \frac{1 \times 8.314 \times 300}{1000} \\ Δ E = - 69.81 kJ / mole \\ \begin{aligned} \begin{aligned} C_{2} H_{4} + HCl ⟶ C_{2} H_{5} Cl \end{aligned} \\ n = 3.5 3 \end{aligned} \\ (LR) \\ Δ E = - 69.81 \times 3 = - 209.43 \end{array}$