Question

Ethyl chloride $\left(C_2{H}_{5}Cl\right)$, is prepared by reaction of ethylene with hydrogen chloride :

$C_2{H}_4\left(g\right)+HCl\left(g\right)⟶C_2{H}_{5}Cl\left(g\right);△H=-72.3kJ/mol$
What is the value of $△U$ (in kJ), if $70g$ of ethylene and $73g$ of $HCL$ are allowed to react at $300K$?

• A. $-69.8$
• B. $-180.75$
• C. $-174.5$
• D. $-139.6$

Answer 1

The correct option is D $-139.6$

Mass of ethylene $=70g$

Molar mass of ethylene $=28g$

$\therefore$ No. of moles of ethylene $=\frac{70}{28}=2.5\text{moles}$

Mass of $HCl=73g$

Molecular mass of $HCl=36.5g$

No. of moles of $HCl=\frac{73}{36.5}=2\text{moles}$

Since no. of moles of $HCl$ are less than that of ethylene.

Hence $HCl$ will be limiting reagent.

${C_2{H}_4}_{\left(g\right)}+{HCl}_{\left(g\right)}⟶{C_2{H}_{5}Cl}_{\left(g\right)}$

From the reaction-

$1$ mole of $HCl$ reacts with $1$ mole of $C_2{H}_4$ and gives $1$ moles of $C_2{H}_{5}Cl$.

$\therefore 2$ moles of $HCl$ will react with $2$ moles of $C_2{H}_4$ and will give $2$ mole of $C_2{H}_{5}Cl$.

Given that $\Delta H=-72.3KJ/mol=-72300KJ/mol$

Given $\Delta H$ is for $1$ mole of product.

For $2$ moles,

$\Delta H=2\times (-72300)=144600KJ/mol$

$\Delta n_g=n_P-n_R=(2-(2+2))=-2$

As we know that,

$\Delta H=\Delta U+\Delta n_{g}RT$

$\Rightarrow \Delta U=\Delta H-\Delta n_{g}RT$

$\Delta U=-144600-(-2\times 8.314\times 300)=-139.61KJ/mol$.