Question

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

$C_2{H}_4\left(g\right)+HCl\left(g\right)\to C_2{H}_{5}Cl\left(g\right)\Delta H=-72.3kJ$. What is the value of $\Delta E$ (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

No. of moles $C_2{H}_4=\frac{70}{28}=2.5$, No. of moles of $HCl$ (limiting Reagent) = $\frac{73}{36.5}=2$

$\Delta H=\Delta E+\Delta n_{g}RT;$

$-72.3=\Delta E+\frac{(-1\times 8.314\times 300)}{1000}$

$\Delta E=-69.80;$ for two moles $\Delta E=-69.80\times 2$

$\Rightarrow -139.6kJ$