Chap14-Part3

# Chap14-Part3 - LAST CLASS ...

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Unformatted text preview: LAST CLASS  ‐ Rate of a reac/on  ‐ Rate constant of a reac/on  ‐ Order of a reac/on  1  First Order Reac/on  k A→B First order reac2on is one whose rate depends on the concentra2on of  a single  reactant raised to the ﬁrst power  Rate = ‐Δ[A]/Δt = k[A]  Δ[A]/[A] = ‐kΔt  (diﬀeren2al rate law)  Integra2on on both sides,  ln[A]t‐ ln[A]0 =‐kt (Integrated rate law)   The concentra2on of A as a func2on of 2me   [ A]t = [ A]0 exp(−kt ) or ln[ At ] = ln[ A 0] − kt Recall that k has units of 2me‐1.  € 2  Problem: The decomposi2on of a certain insec2cide in water follows ﬁrst order  kine2cs with a rate constant of 1.45 yr‐1. If the concentra2on of the insec2cide is  5.0X10‐7g/cm3, (a) what is its concentra2on aRer  1year (b) how long will it take for  the concentra2on of insec2cide to decrease to 3.0X10‐7g/cm3   3  Problem: The decomposi2on of a certain insec2cide in water follows ﬁrst order  kine2cs with a rate constant of 1.45 yr‐1. If the concentra2on of the insec2cide is  5.0X10‐7g/cm3, (a) what is its concentra2on aRer  1year (b) how long will it take for  the concentra2on of insec2cide to decrease to 3.0X10‐7g/cm3   (a) k = 1.45 yr‐1  [A]0 = 5X10‐7 g/cm3  ln[A]t=1 yr = ‐ kt+ ln[A]0       = ‐ (1.45 yr‐1) (1 yr) + ln (5X10‐7)       = ‐1.45+ (‐14.51) = ‐15.96   [A]t=1yr = exp(‐15.96) = 1.2X10‐7 g/cm3    4  Problem: The decomposi2on of a certain insec2cide in water follows ﬁrst order  kine2cs with a rate constant of 1.45 yr‐1. If the concentra2on of the insec2cide is  5.0X10‐7g/cm3, (a) what is its concentra2on aRer  1year (b) how long will it take for  the concentra2on of insec2cide to decrease to 3.0X10‐7g/cm3   (a) k = 1.45 yr‐1  (b) k = 1.45 yr‐1  [A]0 = 5X10‐7 g/cm3  [A]0 = 5X10‐7 g/cm3  ln[A]t=1 yr = ‐ kt+ ln[A]0  ln[A]t = ‐ kt+ ln[A]0       = ‐ (1.45 yr‐1) (1 yr) + ln (5X10‐7)  ln(3X10‐7)= ‐ (1.45 yr‐1) (t yr) + ln (5X10‐7)       = ‐1.45+ (‐14.51) = ‐15.96       t= 0.35 yr   [A]t=1yr = exp(‐15.96) = 1.2X10‐7 g/cm3    5  First Order Irreversible Reac/on:  Plots  1.0 0.8 0.6 B(t)/A(0) B(t)/ A(0)  .4 0 0.2 0.0 0 1 2 3 4 kt Half Life:  Half life t ½ is 2me required for the concentra2on of a  reactant to reach one‐half of its ini2al value   [A]t ½  = ½ [A]0   ln {[A]t/[A]0} = ‐kt    ln {1/2} = ‐kt ½   t ½ = 0.693/k :  The half life does not depend on the ini2al  concentra2on in ﬁrst order reac2on  6  First Order Irreversible Reac/on:  Plots  ln[A]t= ‐kt + ln[A]0 (Integrated rate law)  Y        = mX + C   Slope= ‐k ( k is posi2ve)  [A]  ln[A]  t  t  7  Example: First Order Reac/on  CH3NC  Pressure is a  measurement of  the concentra2on  CH3CN  Gas‐phase isomeriza2on  Slope = ‐5.1X10‐5 s‐1  k= 5.1X10‐5 s‐1   8  Half‐Time  ln [CH3NC]t = ‐kt + ln [CH3NC]0  The half life does not depend on the ini2al concentra2on in a ﬁrst order  9  reac2on  First‐Order Reac/on:  Example  Americium‐241 is used in smoke detectors.  It has a ﬁrst‐order rate constant for  radioac2ve decay of k = 1.6 x 10‐3 yr‐1.    (a) What is the half‐life?  (b) How much of a 1 mg sample remains aRer three half‐lives?  10  First‐Order Reac/on:  Example  Americium‐241 is used in smoke detectors.  It has a ﬁrst‐order rate constant for  radioac2ve decay of k = 1.6 x 10‐3 yr‐1.    (a) What is the half‐life?  t ½ = 0.693/k = 0.693/ 1.6X10‐3 yr‐1 = 433 yrs   (b) How much of a 1 mg sample remains aRer three half‐lives?  11  First‐Order Reac/on:  Example  Americium‐241 is used in smoke detectors.  It has a ﬁrst‐order rate constant for  radioac2ve decay of k = 1.6 x 10‐3 yr‐1.    (a) What is the half‐life?  t ½ = 0.693/k = 0.693/ 1.6X10‐3 yr‐1 = 433 yrs   (b) How much of a 1 mg sample remains aRer three half‐lives?  1) 433 yrs  2) 433 yrs  3) 433 yrs   1 mg  ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐> 0.5 mg ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐> 0.25mg ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐> 0.125mg  12  Second Order Reac/on  Second order reac2on is one whose rate depends on the concentra2on of  a  single reactant raised to the second power  k 2A → B Or         A+ B  Products  Rate = ‐Δ[A]/Δt = k[A]2  Δ[A]/[A]2 = ‐kΔt  (diﬀeren2al rate law)  Integra2on on both sides, 1/[A]t‐ 1/[A]0 = kt (Integrated rate law)   The concentra2on of A as a func2on of 2me is  [ A]t = [ A]0 1 + kt[ A]0 or 1 1 = + kt [ A]t [ A]0 Recall that k has units of concentra2on‐12me‐1.  € 13  Second Order Irreversible Reac/on:  Plots  1.0 half‐2me for reac2on:  0.8 0.6 B(t)/ A(0)  B(t)/A(0) [ A]0 [ A]t = 1 + kt[ A]0 [ A]0 1 / 2[ A]0 = ⇒ (1 + kt1 / 2[ A]0) = 2 1 + kt1/ 2[ A]0 0.4 0.2 0.0 0 1 2 3 kA(0)t t ½ =  1/ k[A]0  The half life depends on the ini2al concentra2on in a second order  14  reac2on  4 Second Order Irreversible Reac/on:  Plots  1/[A]t = kt + 1/[A]0 (Integrated rate law)  Y        = mX + C   1/[A]  [A]  Slope= k  t  t  15  Second‐Order Processes  The decomposi2on of NO2 at 300°C is described by the  equa2on  1  NO2 (g)  NO (g) +      O2 (g)  2  and yields data comparable to this:  Time (s) [NO2], M 0.0 0.01000 50.0 0.00787 100.0 0.00649 200.0 0.00481 300.0 0.00380 16  Second‐Order Processes  •  Plolng ln [NO2] vs. t yields the  graph at the right.  •  The plot is not a straight line,  so the process is not ﬁrst‐ order in [A].  Time (s) [NO2], M ln [NO2] 0.0 0.01000 −4.610 50.0 0.00787 −4.845 100.0 0.00649 −5.038 200.0 0.00481 −5.337 300.0 0.00380 −5.573 17  Second‐Order Processes  •  Time (s) [NO2], M 1/[NO2] 0.0 0.01000 100 50.0 0.00787 127 100.0 0.00649 0.00481 0.00380 263 •  Because this is a  straight line, the  process is second‐ order in [A].  208 300.0 however, gives this  plot.  154 200.0 1  Graphing         vs. t,  [NO2]  18  Sucrose (a disaccharide) reacts in acidic solu2ons to form two  monosaccharides (glucose & fructose).  At 23oC in 0.5 M HCl,   the following data were obtained for the disappearance of sucrose:  2me (min)    0    39    80    140  210            sucrose (M)   0.316   0.274   0.238   0.190   0.146  Is the reac2on ﬁrst or second order with respect to sucrose?  19  Sucrose (a disaccharide) reacts in acidic solu2ons to form two  monosaccharides (glucose & fructose).  At 23oC in 0.5 M HCl,   the following data were obtained for the disappearance of sucrose:  2me (min)    0    39    80    140  210            sucrose (M)   ln[sucrose]    1/[sucrose]   0.316  0.274  0.238  0.190  0.146                      ‐1.152    ‐1.295    ‐1.435    ‐1.661    ‐1924              3.16    3.65    4.20   5.26   6.85          Is the reac2on ﬁrst or second order with respect to sucrose? First order  20  Temperature Dependence of Reac/on Rates  Rate constants k increase with temperature.  Example:  A chemiluminescent reac2on.  At the higher temperature the reac2on   proceeds faster.    21  Temperature Dependence of Reac/on Rates  CH3NC  CH3CN  gas‐phase isomeriza2on  22  A ﬁrst order reac2on is examined  for three diﬀerent condi2ons.  (a) Which two lines represent data  obtained at the same temperature?  (b) Which two lines represent data  obtained with the same ini2al condi2on?  (c) Which line represents data taken  at the lower temperature?  23  A ﬁrst order reac2on is examined  for three diﬀerent condi2ons.  (a) Which two lines represent data  obtained at the same temperature?  Line 1 and 2: Same slope  Started at 2 diﬀerent ini2al concentra2ons  (b) Which two lines represent data  obtained with the same ini2al condi2on?  (c) Which line represents data taken  at the lower temperature?  24  A ﬁrst order reac2on is examined  for three diﬀerent condi2ons.  (a) Which two lines represent data  obtained at the same temperature?  Line 1 and 2: Same slope  Started at 2 diﬀerent ini2al concentra2ons  (b) Which two lines represent data  obtained with the same ini2al condi2on?  Line 2 and 3: Same ini2al concentra2on  But 2 diﬀerent temperatures  (c) Which line represents data taken  at the lower temperature?  25  A ﬁrst order reac2on is examined  for three diﬀerent condi2ons.  (a) Which two lines represent data  obtained at the same temperature?  Line 1 and 2: Same slope  Started at 2 diﬀerent ini2al concentra2ons  (b) Which two lines represent data  obtained with the same ini2al condi2on?  Line 2 and 3: Same ini2al concentra2on  But 2 diﬀerent temperatures  (c) Which line represents data taken  at the lower temperature?  Line 3: Smaller slope  26  ...
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