{"id":136,"date":"2020-06-24T13:32:52","date_gmt":"2020-06-24T13:32:52","guid":{"rendered":"https:\/\/libraryresources.nse.org.ng\/chbe220\/chapter\/kinetic-thermodynamic-con\/"},"modified":"2026-03-16T13:50:13","modified_gmt":"2026-03-16T13:50:13","slug":"kinetic-thermodynamic-con","status":"publish","type":"chapter","link":"https:\/\/libraryresources.nse.org.ng\/chbe220\/chapter\/kinetic-thermodynamic-con\/","title":{"raw":"Kinetic & Thermodynamic Control","rendered":"Kinetic & Thermodynamic Control"},"content":{"raw":"
\n
\n
\n

Learning Objectives<\/p>\n\n<\/header>\n

\n\nBy the end of this section, you should be able to:\n\nUnderstand<\/strong> kinetic and thermodynamic control<\/span>\n\n<\/div>\n<\/div>\n \n\nReactants can sometimes give rise to a variety of products.<\/span>\n\n<\/div>\n
\n
\n\nConsider the nitration of nitrobenzene:\n\n\"\"\n\n<\/div>\n<\/div>\n<\/div>\n
\n
<\/div>\n
\n
\n\nThe relative portion of the products before reaching equilibrium is given by the ratio of the rates of production.\n\n[latex]A+B\u2192P_{1}[\/latex] where [latex]r_{P1}=k_{r1}[A][B][\/latex]\n\n[latex]A+B\u2192P_{2}[\/latex] where [latex]r_{P2}=k_{r2}[A][B][\/latex]\n

Here, before equilibrium:<\/p>\n\n\n\n\n
[latex]\\frac{[P_{2}]}{[P_{1}]}=\\frac{k_{r,2}}{k_{r,1}}[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nThis is called kinetic control<\/strong>, and it is dictated by reaction rates.\nAs opposed to thermodynamic control<\/strong>, which is dictated by reaction equilibrium (after a long time):\n\n \n\nSay we have the system:<\/span>\n\n<\/div>\n<\/div>\n<\/div>\n
\n
\n
\n\n\"\"\n\nIf [latex]k_{e1},k_{e2}\\text{<<}k_{r1},k_{r2}[\/latex]<\/span>\n\nThen at any time before the equilibrium reaction start severely affecting product concentration, the reaction simplifies to:\n\n<\/div>\n<\/div>\n<\/div>\n
\n
\n
\n

\"\"<\/p>\n\n\n\n\n
[latex]\\frac{[P_{1}]}{[P_{2}]}=\\frac{k_{r1}}{k_{r2}}[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nThe reaction is kinetically controlled<\/strong>: the amount of products depends on the rates of reaction<\/strong>.\n\n<\/div>\n<\/div>\n<\/div>\n
\n
<\/div>\n
\n
\n
\n\nProof to show: [latex]\\frac{[P_{1}]}{[P_{2}]}=\\frac{k_{r1}}{k_{r2}}[\/latex]\n\\begin{align*}\nr_{P1}=k_{r1}[A][B]&=\\frac{d[P1]}{dt}\\\\\nr_{P2}=k_{r2}[A][B]&=\\frac{d[P2]}{dt}\n\\end{align*}\n\nSay that both [latex]P_{1}[\/latex], [latex]P_{2}[\/latex] start at a concentration of 0. We can express the change in concentration for [latex]P_{1}[\/latex] and [latex]P_{2}[\/latex] at any time before [A][B] reaches 0. Note that once [A][B] reaches 0, the equilibrium reaction starts to dominate as we no longer have forward reactions that consume A and B to produce P1 and P2.\n

[latex]\\frac{[P1]}{[P2]}=\\frac{\\frac{d[P_{1}]}{dt}}{\\frac{d[P_{2}]}{dt}}=\\frac{k_{r1}[A][B]}{k_{r2}[A][B]}=\\frac{k_{r1}}{k_{r2}}[\/latex]<\/p>\n\n<\/div>\n \n\nIf [latex]k_{e1},k_{e2}>>k_{r1},k_{r2}[\/latex]<\/span>\n\n<\/div>\n<\/div>\n<\/div>\n

\n
<\/div>\n
\n
\n\nThen at any given time, this reaction simplifies to:\n

\"\"<\/p>\n\n\n\n\n
[latex]\\frac{[P_{1}]}{[P_{2}]}=\\frac{k_{e1}}{k_{e2}}[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nThe reaction is thermodynamically controlled<\/strong> : the amount of products depends on the equilibrium state<\/strong>.\n\n<\/div>\n<\/div>\n<\/div>\n
\n
\n
\n\nProof to show: [latex]\\frac{[P_{1}]}{[P_{2}]}=\\frac{k_{e1}}{k_{e2}}[\/latex]:\nAt equilibrium:\n\nForward reaction rate: [latex]\\frac{d[P2]}{dt}=k_{e2}[P1][\/latex]\n\nReverse reaction rate: [latex]\\frac{d[P1]}{dt}=k_{e1}[P2][\/latex]\n\nAt equilibrium, the forward and reverse reaction rates are equal:\n\n\\begin{align*}\nk_{e2}[P1] & = k_{e1}[P2]\\\\\n\\frac{[P_{1}]}{[P_{2}]} & = \\frac{k_{e1}}{k_{e2}}\n\\end{align*}\n\n<\/div>\n \n\n<\/div>\n
\n
\n\n \n\n<\/div>\n<\/div>\n<\/div>","rendered":"
\n
\n
\n
\n

Learning Objectives<\/p>\n<\/header>\n

\n

By the end of this section, you should be able to:<\/p>\n

Understand<\/strong> kinetic and thermodynamic control<\/span><\/p>\n<\/div>\n<\/div>\n

 <\/p>\n

Reactants can sometimes give rise to a variety of products.<\/span><\/p>\n<\/div>\n

\n
\n

Consider the nitration of nitrobenzene:<\/p>\n

\"\"<\/p>\n<\/div>\n<\/div>\n<\/div>\n

\n
<\/div>\n
\n
\n

The relative portion of the products before reaching equilibrium is given by the ratio of the rates of production.<\/p>\n

[latex]A+B\u2192P_{1}[\/latex] where [latex]r_{P1}=k_{r1}[A][B][\/latex]<\/p>\n

[latex]A+B\u2192P_{2}[\/latex] where [latex]r_{P2}=k_{r2}[A][B][\/latex]<\/p>\n

Here, before equilibrium:<\/p>\n\n\n\n
[latex]\\frac{[P_{2}]}{[P_{1}]}=\\frac{k_{r,2}}{k_{r,1}}[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

This is called kinetic control<\/strong>, and it is dictated by reaction rates.
\nAs opposed to thermodynamic control<\/strong>, which is dictated by reaction equilibrium (after a long time):<\/p>\n

 <\/p>\n

Say we have the system:<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n

\n
\n
\n

\"\"<\/p>\n

If [latex]k_{e1},k_{e2}\\text{<<}k_{r1},k_{r2}[\/latex]<\/span><\/p>\n

Then at any time before the equilibrium reaction start severely affecting product concentration, the reaction simplifies to:<\/p>\n<\/div>\n<\/div>\n<\/div>\n

\n
\n
\n

\"\"<\/p>\n\n\n\n
[latex]\\frac{[P_{1}]}{[P_{2}]}=\\frac{k_{r1}}{k_{r2}}[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The reaction is kinetically controlled<\/strong>: the amount of products depends on the rates of reaction<\/strong>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

\n
<\/div>\n
\n
\n
\n

Proof to show: [latex]\\frac{[P_{1}]}{[P_{2}]}=\\frac{k_{r1}}{k_{r2}}[\/latex]
\n\\begin{align*}
\nr_{P1}=k_{r1}[A][B]&=\\frac{d[P1]}{dt}\\\\
\nr_{P2}=k_{r2}[A][B]&=\\frac{d[P2]}{dt}
\n\\end{align*}<\/p>\n

Say that both [latex]P_{1}[\/latex], [latex]P_{2}[\/latex] start at a concentration of 0. We can express the change in concentration for [latex]P_{1}[\/latex] and [latex]P_{2}[\/latex] at any time before [A][B] reaches 0. Note that once [A][B] reaches 0, the equilibrium reaction starts to dominate as we no longer have forward reactions that consume A and B to produce P1 and P2.<\/p>\n

[latex]\\frac{[P1]}{[P2]}=\\frac{\\frac{d[P_{1}]}{dt}}{\\frac{d[P_{2}]}{dt}}=\\frac{k_{r1}[A][B]}{k_{r2}[A][B]}=\\frac{k_{r1}}{k_{r2}}[\/latex]<\/p>\n<\/div>\n

 <\/p>\n

If [latex]k_{e1},k_{e2}>>k_{r1},k_{r2}[\/latex]<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n

\n
<\/div>\n
\n
\n

Then at any given time, this reaction simplifies to:<\/p>\n

\"\"<\/p>\n\n\n\n
[latex]\\frac{[P_{1}]}{[P_{2}]}=\\frac{k_{e1}}{k_{e2}}[\/latex]<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The reaction is thermodynamically controlled<\/strong> : the amount of products depends on the equilibrium state<\/strong>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

\n
\n
\n

Proof to show: [latex]\\frac{[P_{1}]}{[P_{2}]}=\\frac{k_{e1}}{k_{e2}}[\/latex]:
\nAt equilibrium:<\/p>\n

Forward reaction rate: [latex]\\frac{d[P2]}{dt}=k_{e2}[P1][\/latex]<\/p>\n

Reverse reaction rate: [latex]\\frac{d[P1]}{dt}=k_{e1}[P2][\/latex]<\/p>\n

At equilibrium, the forward and reverse reaction rates are equal:<\/p>\n

\\begin{align*}
\nk_{e2}[P1] & = k_{e1}[P2]\\\\
\n\\frac{[P_{1}]}{[P_{2}]} & = \\frac{k_{e1}}{k_{e2}}
\n\\end{align*}<\/p>\n<\/div>\n

 <\/p>\n<\/div>\n

\n
\n

 <\/p>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":1,"menu_order":9,"template":"","meta":{"pb_show_title":"","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-136","chapter","type-chapter","status-publish","hentry"],"part":97,"_links":{"self":[{"href":"https:\/\/libraryresources.nse.org.ng\/chbe220\/wp-json\/pressbooks\/v2\/chapters\/136","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/libraryresources.nse.org.ng\/chbe220\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/libraryresources.nse.org.ng\/chbe220\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/chbe220\/wp-json\/wp\/v2\/users\/1"}],"version-history":[{"count":1,"href":"https:\/\/libraryresources.nse.org.ng\/chbe220\/wp-json\/pressbooks\/v2\/chapters\/136\/revisions"}],"predecessor-version":[{"id":137,"href":"https:\/\/libraryresources.nse.org.ng\/chbe220\/wp-json\/pressbooks\/v2\/chapters\/136\/revisions\/137"}],"part":[{"href":"https:\/\/libraryresources.nse.org.ng\/chbe220\/wp-json\/pressbooks\/v2\/parts\/97"}],"metadata":[{"href":"https:\/\/libraryresources.nse.org.ng\/chbe220\/wp-json\/pressbooks\/v2\/chapters\/136\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/libraryresources.nse.org.ng\/chbe220\/wp-json\/wp\/v2\/media?parent=136"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/chbe220\/wp-json\/pressbooks\/v2\/chapter-type?post=136"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/chbe220\/wp-json\/wp\/v2\/contributor?post=136"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/libraryresources.nse.org.ng\/chbe220\/wp-json\/wp\/v2\/license?post=136"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}