Question

Ethyl chloride ($C_2{H}_{5}Cl$), is prepared by reaction of ethylene with hydrogen chloride:
$C_2{H}_4+HCl\left(g\right)\to C_2{H}_{5}Cl\left(g\right);△H=-72.3kJ$

What is the value of $△E$ (in kJ), if 70 g of ethylene and 73g of HCl are allowed to react at 300 K.

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

Answer 1

The correct option is D -139.6

Solution:-

Weight of ethylene $=70g$

Mol. wt. of ethylene $=28g$

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

Weight of $HCl=73g$

Mol. wt. of $HCl=36.5g$

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

$HCl$ is limiting reagent.

Now,

${C_2{H}_4}_{\left(g\right)}+{HCl}_{\left(g\right)}⟶{C_2{H}_{5}Cl}_{\left(g\right)}\Delta H=-72.3kJ$

From the above reaction,

$1$ mole of $HCl$ gives $1$ mole of $C_2{H}_{5}Cl$.

Therefore,

$2$ moles of $HCl$ gives $2$ moles of $C_2{H}_{5}Cl$.

Again from the above reaction,

$\Delta n_g=n_P-n_R=1-(1+1)=-1$

As we know that,

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

at $300K$,

$-72.3=\Delta E+(-1)\times 8.34\times {10}^{-3}\times 300$

$\Delta E=-72.3+2.5=-69.8kJ$

Now,

For $1$ mole of $C_2{H}_{5}Cl$-

$\Delta E=-69.8kJ$

Therefore,

For $2$ mole of $C_2{H}_{5}Cl$-

$\Delta E=2\times (-69.8)=-139.6kJ$