Question

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

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

Answer 1

The correct option is C $-209.41$

$△n_g=1-2=-1△U=△H-△n_{g}RT=△H+RT=-72.3+8.314\times 300\times {10}^{-3}=-69.8kJ/mol$

Weight of ethylene =98g
Mol. wt. of ethylene =28g
No. of moles of ethylene =98/28​=3.5 mol

Weight of HCl=109.5g
Mol. wt. of HCl=36.5g
No. of moles of HCl=109.5/36.5​=3 mol

HCl is limiting reagent.
So, for 3 moles we will get
$△U=-69.8\times 3kJ=-209.42kJ$