Question

Ethanol, C₂H₅OH,is produced industrially from ethylene, C₂H₄, by the following sequence of reactions:

\begin{aligned}{\rm{3}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{+2}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\to{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{HS}}{{\rm{O}}_{\rm{4}}}{\rm{+(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{{\rm{)}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}\\{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{HS}}{{\rm{O}}_{\rm{4}}}{\rm{ +(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{{\rm{)}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{+3}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\to{\rm{3}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH + 2}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}} \end{aligned}

What volume of ethylene at STP is required to produce 1000 metric tons (1000kg) of ethanol if the overall yield of ethanol is 90.1%?

Short answer

The volume of ethylene is obtained as 5.4×10⁵ L.

Step-by-step solution

Define Gas

A gas is made up of particles with no definite volume or structure

Evaluating the mass of the ethanol produced

The mass of the ethanol produced is:

1000kg = 1×10⁶ g

The overall yield of ethanol is 90.1%.

The mass of the overall liquid produced is:

\(\frac{{{\rm{90}}{\rm{.1 \% }}}}{{100}}{\rm{ \times1\times1}}{{\rm{0}}^{\rm{6}}}{\rm{ g = 1}}{\rm{.1099 \times1}}{{\rm{0}}^{\rm{6}}}{\rm{ g}}\)

So,the mass of the ethanol produced is

\({\rm{1}}{\rm{.1099\times1}}{{\rm{0}}^{\rm{6}}}{\rm{ g}}\).

Evaluating the volume

The following conversions must be made to get from the mass of ethanol to the volume of ethylene:

\({\rm{Mass of }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH }} \to {\rm{ Moles of }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH }} \to {\rm{ Mass of }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{ }} \to {\rm{ Volume of }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\)

So, the first two conversions are:

\({\rm{1}}{\rm{.1099 \times 1}}{{\rm{0}}^{\rm{6}}}{\rm{ g \times }}\frac{{{\rm{1 mol }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}}}{{{\rm{46}}{\rm{.07 }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}}}{\rm{ \times }}\frac{{{\rm{1 mol }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}}}{{{\rm{1 mol }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{OH}}}}{\rm{ = 2}}{\rm{.41 \times 1}}{{\rm{0}}^{\rm{4}}}{\rm{ mol }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\)

Then, using the ideal gas law as:

At, \(\begin{array}{c}{\rm{STP, T = 273 K}}\\{\rm{P = 1 atm}}\end{array}\)

\(\begin{array}{c}{{\rm{V}}_{{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}}}{\rm{ = }}{\left( {\frac{{{\rm{nRT}}}}{{\rm{P}}}} \right)_{{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}}}\\{\rm{ = }}\frac{{{\rm{2}}{\rm{.41 \times 1}}{{\rm{0}}^{\rm{4}}}{\rm{\;mol \times 0}}{\rm{.08206\;L\;atm\;mo}}{{\rm{l}}^{{\rm{ - 1}}}}{\rm{\;}}{{\rm{K}}^{{\rm{ - 1}}}}{\rm{ \times 273\;K}}}}{{{\rm{1\;atm}}}}\\{\rm{ = 5}}{\rm{.4 \times 1}}{{\rm{0}}^{\rm{5}}}{\rm{\;L}}\end{array}\)

Therefore, the volume is: \({\rm{5}}{\rm{.4 \times 1}}{{\rm{0}}^{\rm{5}}}{\rm{\;L}}\).