{"id":1334,"date":"2020-06-23T16:59:43","date_gmt":"2020-06-23T20:59:43","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/chbe220\/?post_type=chapter&p=1334"},"modified":"2020-08-13T12:12:46","modified_gmt":"2020-08-13T16:12:46","slug":"integrated-raw-laws","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/chbe220\/chapter\/integrated-raw-laws\/","title":{"raw":"Integrated Rate Laws","rendered":"Integrated Rate Laws"},"content":{"raw":"
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Learning Objectives<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nBy the end of this section, you should be able to:\r\n\r\nDetermine<\/strong> the reaction order and rate constant from kinetic data using the linearized form of integrated rate laws<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n

Integrated Rate Laws<\/h2>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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\r\n\r\nRate laws are differential equations that can be integrated to find how the concentrations of reactants and products change with time. We can imagine a fairly complex system with multiple reactions described by\r\n

[latex]\ud835\udc5f_{1}=\ud835\udc58_{\ud835\udc5f1}\u2217\ud835\udc53(\ud835\udc34,\ud835\udc35,...), \ud835\udc5f_{2}=\ud835\udc58_{\ud835\udc5f2}\u2217\ud835\udc53(\ud835\udc34,\ud835\udc35,...), \ud835\udc5f_{3}, \ud835\udc5f_{4}...[\/latex]<\/p>\r\nThe concentration of all components can be described by equations such as [latex]\\frac{d[A]}{dt}=\u2212r_{1}\u22122r_{2}...[\/latex] We need to integrate these to find concentrations at a given time.\r\n\r\nAny of these rate laws can be solved numerically (you\u2019ll learn about this in CHBE 230), but some simpler cases can also be solved analytically.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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Zeroth Order Reactions<\/h2>\r\n

A zeroth-order reaction is one whose rate is independent of concentration[latex]^{[1]}[\/latex]; Say we have a reaction:<\/p>\r\n

[latex]A \u2192 B[\/latex]<\/p>\r\nIt's differential rate law would be represented as:\r\n

[latex]r = -\\frac{d[A]}{dt} =k_{r}[\/latex]<\/p>\r\nIntegrating from t=0, when the system has a concentration of A as [latex][A]_{0}[\/latex], to some time t, when the system has a concentration represented by [latex][A][\/latex]\r\n

[latex]\\int_{[A]_{0}}^{[A]} \\mathrm d[A] = \\int_{t_{0}=0}^{t} \\mathrm -k_{r}[\/latex]<\/p>\r\n\r\n\r\n\r\n\r\n
[latex}[A]-[A]_{0}=-k_{r}*t[\/latex] or [latex][A]=[A]_{0}-k_{r}*t[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

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First-Order Reactions<\/h2>\r\nUsing a similar approach to the zeroth order reactions:\r\n

[latex]r=-\\frac{d[A]}{dt}=k_{r}[A][\/latex]<\/p>\r\n

[latex]\\int_{[A]_{0}}^{[A]} \\frac{d[A]}{[A]} = \\int_{t_{0}=0}^{t} \\mathrm-k_{r}[\/latex]<\/p>\r\n\r\n\r\n\r\n\r\n
[latex]ln[A]-ln[A]_{0}=-k_{r}*t[\/latex] or [latex][A]=[A]_{0}e^{-k_{r}*t}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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Example: Half-life for First-order Reactions<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nBased on the equation we derived, what is an expression for the time taken for half of A to be consumed (half-life)?\r\n

[latex]ln[A]-ln[A]_{0}=-k_{r}*t[\/latex]<\/p>\r\n\\begin{align*}\r\nk_{r}* t_{1\/2} & = -ln \\frac{[A]}{[A]_{0}} \\\\\r\n& = -ln \\frac{\\frac{1}{2}[A]_{0}}{[A]_{0}} \\\\\r\n& =-ln(\\frac{1}{2}) \\\\\r\n& = ln(2)\r\n\\end{align*}\r\n

[latex]t_{1\/2}=\\frac{ln(2)}{k_{r}}[\/latex]<\/p>\r\nNote that this half-life does not depend on [A].\r\n\r\nPractice<\/strong>: use [latex][A]=[A]_{0}e^{-k_{r}*t}[\/latex], you should get the same result.\r\n\r\nSolution<\/strong>\r\n\r\n\\begin{align*}\r\n[A] & = [A]_{0}*e^{-k_{r}t_{1\/2}} \\\\\r\n\\frac{[A]}{[A]_{0}}& = e^{-k_{r}t_{1\/2}} \\\\\r\n\\frac{1}{2} & = e^{-k_{r}t_{1\/2}} \\\\\r\nln(\\frac{1}{2}) & = -k_{r}*t_{1\/2}\r\n\\end{align*}\r\n

[latex]t_{1\/2} =-\\frac{ln(\\frac{1}{2})}{k_{r}} = \\frac{ln(2)}{k_{r}}[\/latex]<\/p>\r\n \r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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Second-Order Reactions<\/h2>\r\n

[latex]r=-\\frac{d[A]}{dt}=k_{r}[A]^2[\/latex]<\/p>\r\n

[latex]\\int_{[A]_{0}}^{[A]} \\frac{d[A]}{[A]^2} = \\int_{t_{0}=0}^{t} \\mathrm-k_{r}[\/latex]<\/p>\r\n\r\n\r\n\r\n\r\n
[latex]\\frac{1}{[A]}-\\frac{1}{[A]_{0}}=k_{r}*t[\/latex] or [latex][A]=\\frac{[A]_{0}}{1+k_{r}*t*[A]_{0}}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nUsing this to find half life, we find that: [latex]t_{1\/2}=\\frac{1}{k_{r}*[A]_{0}}[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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References<\/h2>\r\n
\r\n\r\n[1] Chemistry LibreTexts. 2020. 12.4 Integrated Rate Laws.<\/i> [online] Available at: <https:\/\/chem.libretexts.org\/Bookshelves\/General_Chemistry\/Book%3A_Chemistry_(OpenSTAX)\/12%3A_Kinetics\/12.4%3A_Integrated_Rate_Laws#:~:text=We%20measure%20values%20for%20the,different%20concentrations%20of%20the%20reactants.&text=Integrated%20rate%20laws%20are%20determined,various%20times%20during%20a%20reaction.<\/a>> [Accessed 24 April 2020].\r\n\r\n<\/div>\r\n<\/div>\r\n \r\n\r\n<\/div>\r\n<\/div>","rendered":"
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Learning Objectives<\/p>\n<\/header>\n

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

Determine<\/strong> the reaction order and rate constant from kinetic data using the linearized form of integrated rate laws<\/span><\/p>\n<\/div>\n<\/div>\n

Integrated Rate Laws<\/h2>\n<\/div>\n<\/div>\n<\/div>\n
\n
\n
\n

Rate laws are differential equations that can be integrated to find how the concentrations of reactants and products change with time. We can imagine a fairly complex system with multiple reactions described by<\/p>\n

[latex]\ud835\udc5f_{1}=\ud835\udc58_{\ud835\udc5f1}\u2217\ud835\udc53(\ud835\udc34,\ud835\udc35,...), \ud835\udc5f_{2}=\ud835\udc58_{\ud835\udc5f2}\u2217\ud835\udc53(\ud835\udc34,\ud835\udc35,...), \ud835\udc5f_{3}, \ud835\udc5f_{4}...[\/latex]<\/p>\n

The concentration of all components can be described by equations such as [latex]\\frac{d[A]}{dt}=\u2212r_{1}\u22122r_{2}...[\/latex] We need to integrate these to find concentrations at a given time.<\/p>\n

Any of these rate laws can be solved numerically (you\u2019ll learn about this in CHBE 230), but some simpler cases can also be solved analytically.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

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\n
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Zeroth Order Reactions<\/h2>\n

A zeroth-order reaction is one whose rate is independent of concentration[latex]^{[1]}[\/latex]; Say we have a reaction:<\/p>\n

[latex]A \u2192 B[\/latex]<\/p>\n

It’s differential rate law would be represented as:<\/p>\n

[latex]r = -\\frac{d[A]}{dt} =k_{r}[\/latex]<\/p>\n

Integrating from t=0, when the system has a concentration of A as [latex][A]_{0}[\/latex], to some time t, when the system has a concentration represented by [latex][A][\/latex]<\/p>\n

[latex]\\int_{[A]_{0}}^{[A]} \\mathrm d[A] = \\int_{t_{0}=0}^{t} \\mathrm -k_{r}[\/latex]<\/p>\n\n\n\n
[latex]-[A]_{0}=-k_{r}*t[\/latex] or [latex][A]=[A]_{0}-k_{r}*t[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

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First-Order Reactions<\/h2>\n

Using a similar approach to the zeroth order reactions:<\/p>\n

[latex]r=-\\frac{d[A]}{dt}=k_{r}[A][\/latex]<\/p>\n

[latex]\\int_{[A]_{0}}^{[A]} \\frac{d[A]}{[A]} = \\int_{t_{0}=0}^{t} \\mathrm-k_{r}[\/latex]<\/p>\n\n\n\n
[latex]ln[A]-ln[A]_{0}=-k_{r}*t[\/latex] or [latex][A]=[A]_{0}e^{-k_{r}*t}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

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Example: Half-life for First-order Reactions<\/p>\n<\/header>\n

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Based on the equation we derived, what is an expression for the time taken for half of A to be consumed (half-life)?<\/p>\n

[latex]ln[A]-ln[A]_{0}=-k_{r}*t[\/latex]<\/p>\n

\\begin{align*}
\nk_{r}* t_{1\/2} & = -ln \\frac{[A]}{[A]_{0}} \\\\
\n& = -ln \\frac{\\frac{1}{2}[A]_{0}}{[A]_{0}} \\\\
\n& =-ln(\\frac{1}{2}) \\\\
\n& = ln(2)
\n\\end{align*}<\/p>\n

[latex]t_{1\/2}=\\frac{ln(2)}{k_{r}}[\/latex]<\/p>\n

Note that this half-life does not depend on [A].<\/p>\n

Practice<\/strong>: use [latex][A]=[A]_{0}e^{-k_{r}*t}[\/latex], you should get the same result.<\/p>\n

Solution<\/strong><\/p>\n

\\begin{align*}
\n[A] & = [A]_{0}*e^{-k_{r}t_{1\/2}} \\\\
\n\\frac{[A]}{[A]_{0}}& = e^{-k_{r}t_{1\/2}} \\\\
\n\\frac{1}{2} & = e^{-k_{r}t_{1\/2}} \\\\
\nln(\\frac{1}{2}) & = -k_{r}*t_{1\/2}
\n\\end{align*}<\/p>\n

[latex]t_{1\/2} =-\\frac{ln(\\frac{1}{2})}{k_{r}} = \\frac{ln(2)}{k_{r}}[\/latex]<\/p>\n

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

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Second-Order Reactions<\/h2>\n

[latex]r=-\\frac{d[A]}{dt}=k_{r}[A]^2[\/latex]<\/p>\n

[latex]\\int_{[A]_{0}}^{[A]} \\frac{d[A]}{[A]^2} = \\int_{t_{0}=0}^{t} \\mathrm-k_{r}[\/latex]<\/p>\n\n\n\n
[latex]\\frac{1}{[A]}-\\frac{1}{[A]_{0}}=k_{r}*t[\/latex] or [latex][A]=\\frac{[A]_{0}}{1+k_{r}*t*[A]_{0}}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Using this to find half life, we find that: [latex]t_{1\/2}=\\frac{1}{k_{r}*[A]_{0}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

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References<\/h2>\n
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[1] Chemistry LibreTexts. 2020. 12.4 Integrated Rate Laws.<\/i> [online] Available at: <https:\/\/chem.libretexts.org\/Bookshelves\/General_Chemistry\/Book%3A_Chemistry_(OpenSTAX)\/12%3A_Kinetics\/12.4%3A_Integrated_Rate_Laws#:~:text=We%20measure%20values%20for%20the,different%20concentrations%20of%20the%20reactants.&text=Integrated%20rate%20laws%20are%20determined,various%20times%20during%20a%20reaction.<\/a>> [Accessed 24 April 2020].<\/p>\n<\/div>\n<\/div>\n

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