A particular reaction, A Product, has a rate that slowsdown as the reaction proc

Question

A particular reaction, A Product, has a rate that slowsdown as the reaction proceeds. The half-life of the reaction isfound to depend on the initial concentration of A. Determinewhether each of the following is likely to be true or false forthis reaction. a. A doubling of the concentration of A doubles the rate ofthe reaction. b.A plot of 1/[A] versus time is linear. c. The half-life of the reaction gets longeras the initial concentration of A increases. d.A plot of the concentration of A versustime has a constant slope. A particular reaction, A Product, has a rate that slowsdown as the reaction proceeds. The half-life of the reaction isfound to depend on the initial concentration of A. Determinewhether each of the following is likely to be true or false forthis reaction. a. A doubling of the concentration of A doubles the rate ofthe reaction. b.A plot of 1/[A] versus time is linear. c. The half-life of the reaction gets longeras the initial concentration of A increases. d.A plot of the concentration of A versustime has a constant slope. c. The half-life of the reaction gets longeras the initial concentration of A increases. d.A plot of the concentration of A versustime has a constant slope. d.A plot of the concentration of A versustime has a constant slope.

Explanation / Answer

a)    doubling concentration of A doubles the rateof the reaction => rate = k*[A] = -d[A]/dt -d[A]/[A] = k*dt -ln([A]/[A0]) = kt when [A] = 0.5[A]0    -ln(0.5) =k*t_half    t_half = ln(2)/ k    => this half life is notdependent on the initial concentration of A. So this is false b) 1/[A] vs t is linear => second order reaction d[A]/dt = -k[A]2 -d[A]/[A]2 = k*dt 1/[A] - 1/[A]0 = k*t when [A] = 0.5[A]0   => 1/[A] = 2/[A]0 2/[A]0 - 1/[A]0 = k*t_half 1/[A]0 = k*t_half t_half = 1/(k*[A]0)      so this istrue c) "rate that slows down as the reaction proceeds" rate = -d[A]/dt =k*[A]n              n> 0         (if n slope of -k => dA/dt = -k [A] - [A0] = -kt t_half = [A0]/2k So this is true; it does depend on the initial concentration

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