Derive integrated rate law expressions for half-, one and a half-, and third-order reactions of the form products where k is the rate constant

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Derive integrated rate law expressions for half-, one and a half-, and third-order reactions of the form products where k is the rate constant. Also, deduce the graphical linear relations, units of rate constant k, and expression for half-life, t1/2 , for all cases. Assume that at t = 0 only A is present with an initial concentration of [A]0. Fill in the table below by following the example given for first-order reaction1. Show all your work. order Integrated Rate Law Graphical Linear Relation Unit of k Half-life 1 ln[A][A]0=−?? ln[A] vs t ⇒ slope = −k , y-intercept = ln[A]0 1? ?12�=ln2? 1/2 3/2 3
Derive integrated rate law expressions for half-, one and a half-, and third-order reactions of the form A ® products where k is the rate constant. Also, deduce the graphical linear relations, units of rate constant k, and expression for half-life, t1/2 , for all cases. Assume that at t = 0 only A is present with an initial concentration of [A]0. Fill in the table below by following the example given for first-order reaction1.

Show all your work.

order Integrated Rate Law Graphical Linear Relation Unit of k Half-life

1 ln[A][A]0=−?? ln[A] vs t ⇒ slope = −k , y-intercept = ln[A]0 1? ?12�=ln2?

1/2

3/2

3

1Atkins’ Physical Chemistry, Atkins and de Paula, 10th Ed., 2014, OUP, pp. 827-828.

The Correct Answer and Explanation is :

To derive the integrated rate laws, graphical linear relations, units of the rate constant ( k ), and half-life expressions for reactions of orders ½, 1½, and 3, we start with the general rate law for a reaction ( A \rightarrow \text{products} ):

[ -\frac{d[A]}{dt} = k[A]^n ]

where ( n ) is the reaction order, ( [A] ) is the concentration of reactant ( A ) at time ( t ), and ( k ) is the rate constant.

Half-Order (n = ½):

  1. Integrated Rate Law: Separate variables and integrate: [ \int [A]^{-1/2} \, d[A] = -k \int dt ] This yields: [ 2[A]^{1/2} = -kt + C ] Applying the initial condition ( [A] = [A]_0 ) at ( t = 0 ): [ 2[A]_0^{1/2} = C ] Thus, the integrated rate law is: [ 2[A]^{1/2} = -kt + 2[A]_0^{1/2} ] Simplifying: [ [A]^{1/2} = [A]_0^{1/2} – \frac{kt}{2} ]
  2. Graphical Linear Relation: Plotting ( [A]^{1/2} ) versus ( t ) yields a straight line with:
  • Slope: ( -\frac{k}{2} )
  • Y-intercept: ( [A]_0^{1/2} )
  1. Units of ( k ): From the rate law ( k[A]^{1/2} ), the units of ( k ) are: [ \text{(concentration/time)} \times (\text{concentration})^{-1/2} = \text{concentration}^{1/2} \, \text{time}^{-1} ]
  2. Half-Life (( t_{1/2} )): The half-life is the time at which ( [A] = \frac{[A]_0}{2} ). Substituting into the integrated rate law: [ \left( \frac{[A]0}{2} \right)^{1/2} = [A]_0^{1/2} – \frac{kt{1/2}}{2} ] Solving for ( t_{1/2} ): [ t_{1/2} = \frac{[A]_0^{1/2}}{k} ]

One and a Half Order (n = 3/2):

  1. Integrated Rate Law: Separate variables and integrate: [ \int [A]^{-3/2} \, d[A] = -k \int dt ] This yields: [ -2[A]^{-1/2} = -kt + C ] Applying the initial condition ( [A] = [A]_0 ) at ( t = 0 ): [ -2[A]_0^{-1/2} = C ] Thus, the integrated rate law is: [ -2[A]^{-1/2} = -kt – 2[A]_0^{-1/2} ] Simplifying: [ [A]^{-1/2} = [A]_0^{-1/2} + \frac{kt}{2} ]
  2. Graphical Linear Relation: Plotting ( [A]^{-1/2} ) versus ( t ) yields a straight line with:
  • Slope: ( \frac{k}{2} )
  • Y-intercept: ( [A]_0^{-1/2} )
  1. Units of ( k ): From the rate law ( k[A]^{3/2} ), the units of ( k ) are: [ \text{(concentration/time)} \times (\text{concentration})^{-3/2} = \text{concentration}^{-1/2} \, \text{time}^{-1} ]
  2. Half-Life (( t_{1/2} )): The half-life is the time at which ( [A] = \frac{[A]_0}{2} ). Substituting into the integrated rate law: [ \left( \frac{[A]0}{2} \right)^{-1/2} = [A]_0^{-1/2} + \frac{kt{1/2}}{2} ] Solving for ( t_{1/2} ): [ t_{1/2} = \frac{2(2^{1/2} – 1)}{k[A]_0^{1/2}} ]

Third Order (n = 3):

  1. Integrated Rate Law: Separate variables and integrate: [ \int [A]^{-3} \, d[A] = -k \int dt ] This yields: [ \frac{1}{2}[A]^{-2} = -kt + C ] Applying the initial condition ( [A] = [A]_0 ) at ( t = 0 ): [ \frac{1}{2}[A]_0^{-2} = C ] Thus, the integrated rate law is: [ \frac{1}{2}[A]^{-2} = -kt + \frac{1}{2}[A]_0^{-2} ] Simplifying: [ \frac{1}{[A]^2} = \frac{1}{[A]_0^2} + 2kt ]
  2. Graphical Linear Relation: Plotting ( \frac{1}{[A]^2} ) versus ( t ) yields a straight line with:
  • Slope: ( 2k )
  • Y-intercept: ( \frac{1}{[A]_0^2} )
  1. Units of ( k ): From the rate law ( k[A]^3 ), the units of ( k ) are: [ \text{(concentration/time)} \times (\text{concentration})^{-3} = \text{concentration}^{-2} \, \text{time}^{-1} ]
  2. Half-Life (( t_{1/2} )): The half-life is the time at which ( [A] = \frac{[A]_0}{2} ). Substituting into the integrated rate law: [ \frac{1}{\left( \frac{[A]0}{2} \right)^2} = \frac{1}{[A]_0^2} + 2kt{1/2} ] Solving for ( t_{1/

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