Question

Ethyl chloride $\left( {{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{Cl}}} \right)$, is prepared by the reaction of ethylene with hydrogen chloride:
${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\,{\text{ + }}\,{\text{HCl(g)}}\, \to {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{Cl}}\left( {\text{g}} \right)$ ; $\delta {{H}}\,{\text{ = }}\, - 72.3\,{\text{kJ}}$
What is the value\[{{\delta E}}\] of (in kJ), if $70$ g of ethylene and $73$ g of HCl are allowed to react at $300$ K.

Answer 1

Hint: First we will determine the moles of each reactant. By comparing the mole ratio of both reactants according to the given equation we will determine the limiting reagent. After finding the limiting reagent, by comparing the amount of the limiting reagent with the product according to the given reaction, we will determine the mole of the product. Then we will determine the internal energy change by using the internal energy and enthalpy change relation.

Formula used: \[{{\delta H}}\, = \,\,{{\delta E}}\,\,{\text{ + }}\,\,{{\delta }}{{\text{n}}_{\text{g}}}{\text{RT}}\,\]

Complete step-by-step answer:
We will determine the mole of ethylene as follows:
${\text{mole}}\,{\text{ = }}\,\dfrac{{{\text{mass}}}}{{\,{\text{molar}}\,{\text{mass}}}}$
For ethylene:
Molar mass of ethylene is $28$ g/mol.
On substituting $70$g for mass and $28$ g/mol for molar mass,
${\text{mole}}\,{\text{ = }}\,\dfrac{{70\,{\text{g}}}}{{\,28\,{\text{g/mol}}}}$
${\text{mole}}\,{\text{ = }}\,2.5$
So, the mole of ethylene is $2.5$ mole.

For HCl:
Molar mass of HCl is $36.5$ g/mol.
On substituting $73$g for mass and $36.5$ g/mol for molar mass,
${\text{mole}}\,{\text{ = }}\,\dfrac{{73\,{\text{g}}}}{{\,36.5\,{\text{g/mol}}}}$
${\text{mole}}\,{\text{ = }}\,2$
So, the mole of HCl is $2$ mole.

The reactant which is present in a low amount in the reaction is known as the limiting reagent. Here, HCl is present in low amounts so, HCl is limiting reagent so, we will decide the amount of ethyl chloride on the basis of amount of HCl.

The given reaction is as follows:
${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\,{\text{ + }}\,{\text{HCl(g)}}\, \to {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{Cl}}\left( {\text{g}} \right)$

According to the above reaction, one mole of HCl gives one mole of ethyl chloride so, two moles of HCl will give,
$1$mole HCl = $1$ mole ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{Cl}}$
$2$ mole HCl = $2$mole ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{Cl}}$
So, the amount of product ethyl chloride is $2$mol.

Now, we will determine the \[{{\delta E}}\] as follows:
\[{{\delta H}}\, = \,\,{{\delta E}}\,\,{\text{ + }}\,\,{{\delta }}{{\text{n}}_{\text{g}}}{\text{RT}}\,\]
Where,
\[{{\delta H}}\] is the enthalpy change
\[{{\delta E}}\] is the change in internal energy
\[{{\delta }}{{\text{n}}_{\text{g}}}\] is the change in moles of gaseous species.
R is the gas constant
T is the temperature
\[{{\delta }}{{\text{n}}_{\text{g}}}\] is determined as follows:
\[{{\delta }}{{\text{n}}_{\text{g}}}\,{\text{ = }}\,{{\Sigma }}{{\text{n}}_{\text{g}}}{\text{(product)}} - {{\Sigma }}{{\text{n}}_{\text{g}}}{\text{(reactant)}}\]

According to the given reaction, ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\,{\text{ + }}\,{\text{HCl(g)}}\, \to {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{Cl}}\left( {\text{g}} \right)$ and from above calculation, we know that
${\text{2}}{\text{.5}}\,{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{4}}}\,{\text{ + }}\,2\,{\text{HCl(g)}}\, \to 2{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{Cl}}\left( {\text{g}} \right)$, the moles of HCl(g) is two and moles of ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}{\text{Cl}}\left( {\text{g}} \right)$is also two.
So,
\[{{\delta }}{{\text{n}}_{\text{g}}}\,{\text{ = }}\,{\text{2}} - ({\text{2 + 2)}}\]
\[{{\delta }}{{\text{n}}_{\text{g}}}\,{\text{ = }}\, - 2\]
Now, \[{{\delta H}}\] is given for the \[{{\delta }}{{\text{n}}_{\text{g}}}\,{\text{ = }}\, - 1\] so, for \[{{\delta }}{{\text{n}}_{\text{g}}}\,{\text{ = }}\, - 2\], the \[{{\delta H}}\] will be double so,
\[{{\delta H}}\,{\text{ = }}\, - {\text{72}}{\text{.3}}\, \times \,{\text{2}}\]
\[{{\delta H}}\,{\text{ = }}\, - 144.6\]kJ
On substituting \[ - 2\]mol for \[{{\delta }}{{\text{n}}_{\text{g}}}\], $8.314 \times \,{10^{ - 3}}\,{\text{kJ}}\,\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$ for R, $300$ K for T, and $ - 144.6\,{\text{kJ}}$ for $\delta {\{H}}\,$,
\[{{\delta H}}\, = \,\,{{\delta E}}\,\,{\text{ + }}\,\,{{\delta }}{{\text{n}}_{\text{g}}}{\text{RT}}\,\]
\[ - {\text{144}}{\text{.6}}\,{\text{kJ}} = \,\,{{\delta E}}\,\,{\text{ + }}\,\, - 2\,{\text{mol}} \times 8.314 \times \,{10^{ - 3}}\,{\text{kJ}}\,\,{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}\, \times {\text{300}}\,{\text{K}}\,\]
\[ - {\text{144}}{\text{.6 kJ}}\, = \,\,{{\delta E}}\,\,{\text{ + }}\,\, - 4.9\,{\text{kJ}}\]
\[\,{{\delta E}} = \,\, - {\text{144}}{\text{.6 kJ}}\,\,{\text{ + }}\,4.9\,{\text{kJ}}\]
\[\,{{\delta E}} = \, - {\text{1}}39.7\,{\text{kJ}}\]
So, the value\[{{\delta E}}\] is \[ - {\text{1}}39.7\] kJ, if $70$ g of ethylene and $73$ g of HCl are allowed to react at $300$ K. Therefore, \[ - {\text{1}}39.7\]kJ is the correct answer.

Note: Limiting reagent decides the amount of product because all reactants react in fixed composition as the limiting reagent is in less amount, so it will consume first and other reactant remains in excess. This comparison of the coefficient is known as the stoichiometric comparison. For a stoichiometric comparison, a balanced equation is necessary. If the compounds or molecules are given in place of alphabets it is necessary to balance the given equation first. When the reactants are reacting in a one-one ratio then the reactant which is given in a low amount will be the limiting reagent. When the stoichiometry of reactants differs then we have to determine the limiting reagent because in that case, the one which is given in high amounts can also be the limiting reagent it depends upon the stoichiometry.