Question

Ethyl chloride $\left(C_2{H}_{5}Cl\right)$ is prepared by reaction of ethylene with hydrogen chloride:
$C_2{H}_4\left(g\right)+HCl\left(g\right)\to C_2{H}_{5}Cl\left(g\right)\Delta H=-72.3\text{kJ/mol}$
What is the value of $\Delta E$ (in kJ), if $98\text{g}$ of ethylene and $109.5\text{g}$ of $HCl$ are allowed to react at $300\text{K}$ ?

• A. $-64.81$
• B. $-190.71$
• C. $-209.41$
• D. $-224.38$

Answer 1

The correct option is C $-209.41$

Change in internal enrgy and enthalpy of the reaction is related by,
$\Delta E=\Delta H-\Delta n_{g}RT$
where,
$\Delta n_g=$ Number of moles of gaseous product $-$ Number of moles of gaseous reactant.

In given reaction,
$C_2{H}_4\left(g\right)+HCl\left(g\right)\to C_2{H}_{5}Cl\left(g\right)$

$\Delta n_g=1-2=-1$

$\Delta E=\Delta H-\Delta n_{g}RT=\Delta H-RT=-72.3+8.314\times 300\times {10}^{-3}$
$=-69.806\text{kJ/mol}$

From the data given, we found that HCl is the limiting reagent, thus 3 moles of the $C_2{H}_{5}Cl$ is formed.

So for 3 moles we will get, $\Delta E=-69.806\times 3=-209.418\text{kJ}$