A reacts to form P. A plot of the reciprocal of the concentration A vs. time is

Question

A reacts to form P. A plot of the reciprocal of the concentration A vs. time is a straight line. When the initial concentration of A is 1.0 Times 10^-2 M, its half-life is found to be 20 min. Enter the order of the reaction (example, fifth-order). Enter the number of the right one-line mechanism: A + B rightarrow P A rightarrow P 3A rightarrow P 2A rightarrow P Enter the value of the rate constant for the mechanism in part b) where concentrations are in M units and time in minutes. Use two significant figures and do not enter the units. When the initial concentration of A is 3.0 Times 10^-3 M, what is the half-life in minutes? Enter the value to two significant figures with no units.

Explanation / Answer

Answer – We are given, A reacts and form the P. The plot 1/[A] Vrs time gives straight line and we know when 1/[A] Vrs time show straight line then it is second order of reaction.

Initial concentration [Ao] = 1.0*10-2 M , half-life t ½ = 20 min

a) We know a plot of the 1/[A] Vrs time show straight line then it is second order of reaction.

b) We know given reaction A to P is second order and when the number of the right one-line mechanism there is A to P and A with 2 coefficients, since the one step reaction are elementary reaction and we can consider the mole ratio is the order for the reaction. So when the number of the right one-line mechanism there is A to P is as follow –

4) 2A -----> P

c) We are given half-life t ½ = 20 min and we know this I second order, so the formula for the rate constant for the second order of reaction as follow –

t ½ = 1/ k[Ao]

so, rate constant, k = 1 / t ½ *[Ao]

                               = 1/ 20 min * 1.0*10-2 M

                               = 5.0

d) In this one we are given, [Ao] = 3.0*10-3 M, rate constant calculated

k = 5.0 M-1.min-1

we know the half-life of second order of reaction formula

t ½ = 1/ k[Ao]

      = 1/ 5.0 M-1.min-1 * 3.0*10-3 M

      = 67 min

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