Chemistry - Thermochemistry Question with Solution | TestHub

Given reaction:

$$C_2{H}_4(g)+HCl(g)\to C_2{H}_{5}Cl(g),\Delta H=-72.3kJ/mol$$

Step 1: Find limiting reagent

Moles:

* Ethylene $C_2{H}_4=\frac{70}{28}=2.5mol$

* $HCl=\frac{73}{36.5}=2.0mol$

So, $HCl$ is the limiting reagent, meaning a $2mol$ reaction occurs.

Step 2: Use relation between $\Delta H$ and $\Delta U$

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

Gas moles:

* Reactants = $2$ ($C_2{H}_4+HCl$)

* Products = $1$ ($C_2{H}_{5}Cl$)

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

So, for a $1mol$ reaction:

$$\Delta U=\Delta H-\Delta n_{g}RT=-72.3-(-1)(RT)=-72.3+RT$$

Now, for a $2mol$ reaction:

$$\Delta U_{total}=2(-72.3+RT)$$

At $300K$:

$$RT=8.314\times 10^{-3}\times 300=2.494kJ/mol$$

So:

$$\Delta U=2[-72.3+2.494]=2[-69.806]=-139.6kJ$$

Final Answer:

$$\boxed{-139.6kJ\approx -140kJ}$$