Q26E
Question
Consider the production of \({\bf{100}}\) kg of sodium metal using a current of \({\bf{50}},{\bf{000}}{\rm{ }}{\bf{A}},\) assuming a \({\bf{100}}\% \) yield.
(a) How long will it take to produce the 100 kg of sodium metal?
(b) What volume of chlorine at \({\bf{25}}{\rm{ }}^\circ {\bf{C}}\)and \({\bf{1}}.{\bf{00}}\) atm forms?
Step-by-Step Solution
Verified(a) The time it takes to produce the \(100\;kg\) of sodium metal is \(t = 2.33\;h.\)
(b) The volume of chlorine at \(\;{25^o}C\)and \(1.00\) atm is \(V\left( {C{l_2}} \right) = 539\;{m^3}.\)
The ideal gas law refers to a hypothetical equation of state concerninganideal gas.The Boyles’ law, Charles’ law, and combined gas law give rise to the ideal gas law. It is expressed as:
\({\bf{pV = nRT}}.\)
a) The reaction is:
\(N{a^ + } + {e^ - } \to Na.\)
Use the mass of sodium (m) and calculate the number of moles (n) of sodium.
\(n(Na) = \frac{{m(Na)}}{{M(Na)}}.\)
The values are substituted.
\(\begin{aligned}{l}n(Na) = \frac{{1 \times {{10}^5}{\rm{g}}}}{{22.99{\rm{g}}/{\rm{mol}}}}\\\;\;\;\;\;\;\;\; = 4.35 \times {10^4}{\rm{ mol}}{\rm{.}}\end{aligned}\)
The number of electrons transferred in this reaction is the same as the number of moles of sodium produced.
First, calculate the number of electrons from the number of moles by multiplying the number of moles (n) with the Avogadro's constant\({N_a}\).
\(\begin{aligned}{\underline{\phantom{xx}}}n(Na) = n\left( {{e^ - }} \right)\\N\left( {{e^ - }} \right) = {N_a} \times n\left( {{e^ - }} \right)\\\;\;\;\;\;\;\;\;\; = 6.022 \times {10^{23}}{\rm{mo}}{{\rm{l}}^{ - 1}} \times 4.35 \times {10^4}{\rm{mol}}\\\;\;\;\;\;\;\;\;\; = 2.62 \times {10^{27}}.\end{aligned}\)
One electron has the charge of \(1{e^ - } = 1.60217646 \times {10^{ - 19}}{\rm{C}}\). Use this expression to calculate the transferred charge in this process.
\(\begin{aligned}{lQ = N\left( {{e^ - }} \right) \times 1.602 \times {10^{ - 19}}C\\\;\;\; = 2.62 \times {10^{ - 27}} \times 1.602 \times {10^{ - 19}}C\\\;\;\; = 4.20 \times {10^8}C.\end{aligned}\)
Finally, calculate the time required for the calculated amount of transferred charge with the given current.
If you wish to have some feeling of what amount of time it will be, convert the seconds into hours by dividing the time with the number of seconds that are in anhour:\(1h = 60\;{\rm{min}} \times 60\frac{{\rm{s}}}{{\min }}.\)
\(\begin{aligned}{\underline{\phantom{xx}}}{\bf{Current (I) = }}\frac{{{\bf{Charge (Q)}}}}{{{\bf{Time (t)}}}} \to {\bf{t = }}\frac{{\bf{Q}}}{{\bf{I}}}\\t = \frac{{4.20 \times {{10}^8}C}}{{50,000}}\\ = 8393{\rm{s}} = 2.33\;{\rm{h}}{\rm{.}}\end{aligned}\)
The time it takes to produce the \(100\;kg\) of sodium metal is \(t = 2.33\;h.\)
The reaction is:
\(2C{l^ - } \to {C_2} + 2{e^ - }.\)
The number of moles is evaluated as:
\(\begin{aligned}{\underline{\phantom{xx}}}m\left( {{C_2}} \right) = \frac{1}{2} \times m\left( {{e^ - }} \right)\\n\left( {C{l_2}} \right) = \frac{1}{2} \times 4.35 \times {10^4}{\rm{mol}}\\\;\;\;\;\;\;\;\;\; = 2.18 \times {10^4}{\rm{mol}}{\rm{.}}\end{aligned}\)
Use the ideal gas law to calculate the volume.
\({\bf{pV = nRT}} \to {\bf{V = }}\frac{{{\bf{nRT}}}}{{\bf{p}}}.\)
\(\begin{aligned}{\underline{\phantom{xx}}}V\left( {{\rm{C}}{{\rm{l}}_2}} \right) = \frac{{2.18 \times {{10}^4}{\rm{mol}} \times 8.314{\rm{J}}/({\rm{Kmol}}) \times 298{\rm{K}}}}{{{{10}^5}{\rm{Pa}}}}\\\;\;\;\;\;\;\;\;\;\; = 539\;{m^3}.\end{aligned}\)
The volume of chlorine at \(\;{25^o}C\)and \(1.00\) atm is \(V\left( {C{l_2}} \right) = 539\;{m^3}.\)