The integrated rate laws for zero-, first-, and second-order reaction may be arr

Question

The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they resemble the equation for a straight line, y = mx + b. Part A The reactant concentration in a zero-order reaction was 0.100 M after 145 s and 1.50 times 10^-2 M after 330 s. What is the rate constant for this reaction? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash. Part B What was the initial reactant concentration for the reaction described in Part A? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash. Part C The reactant concentration in a first-order reaction was 8.00 times 10^-2 M after 30.0 s and 7.70 times 10^-3 M after 65.0 s. What is the rate constant for this reaction? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash. Part D The reactant concentration in a second-order reaction was 0.690 M after 225 s and 1.30 times 10^-2 M after 895 s. What is the rate constant for this reaction? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash.

Explanation / Answer

Part A :

For a zero order reaction Rate constant , k = ([R]'-[R]) / (t - t')

Where

[R] = concentration at time t = 0.100 M

[R]' = concentration at time t' = 1.50x10-2 M

t = initial time = 145 s

t' = final time = 300 s

Plug the values we get k = ([R]'-[R]) / (t - t')

                                        = ((1.50x10-2 ) - 0.100 ) / (145-300)

                                        = 5.48x10-4 Ms-1

Part B :

For a zero order reaction initial concentration , [R]o = [R] + kt

                                                                                  = 0.100 M + (5.48x10-4 Ms-1 x 145s)

                                                                                  = 0.179 M

Part C :

For a first order reaction rate constant , k = [1/(t2 - t1 )] ln ([R]1/ [R]2)

Where

[R]1 = concentration at time t1 = 8.00x10-2 M

[R]2 = concentration at time t2 = 7.70x10-3 M

t1 = initial time = 30.0 s

t2 = final time = 65.0 s

Plug the values we get k = [1/(65.0-30.0)] ln ((8.00x10-2 )/(7.70x10-3 ))

                                         = 0.670 s-1

Part D :

For a second order reaction,

[(1/[R]2) - (1/[R]1) = k (t2-t1)

Where

[R]1 = concentration at time t1 = 0.690 M

[R]2 = concentration at time t2 = 1.30x10-2 M

t1 = initial time = 225 s

t2 = final time = 896 s

Plug the values we get

[(1/[R]2) - (1/[R]1) = k (t2-t1)

k = 0.112 M-1 s-1

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