Question

2. Plot the concentrations listed below as a function of time. Then, plot any other graphs required to verify that the data is consistent with a second-order reaction. time Concentration (in minutes)(in moles/liter) 0.00 1.0000 1.00 0.8696 2.00 0.7692 3.00 0.6896 4.00 0.6250 5.00 0.5714 6.00 0.5263 7.00 0.4878 8.00 0.4545 9.00 0.4255 10.00 0.40001 (b) Determine the value of the rate constant k. Write an expression for the rate law. (c) Determine the time at which [A] =0.500 mol/L. Determine the time at which [A] = 0.250 mol/L. (d) The half-life of a second-order reaction is not constant. Is this consistent with the times that you calculated in part (c) above? Explain.

Answer 1

For 2nd order reaction 1/[concentration] vs time is straight line with slope = rate constant

b) K = 0.15 for this reaction

F G H K M o N Q P R כ S T Chart Title y = 0.15x+1 1/C 0 1 1 1.149954 2 1.300052 3 1.450116 4 1.6 5 1.750088 6 1.900057 7 2.050021 8 2.20022 9 2.350176 10 2.5 3 2.5 2 15 2 1 0.5 0 0 2 4 8 10 12 12

Expression of rate law =  r = k[A]² = 0.15[A]²

c) 1/[R_t] – 1/[R_o]=kt

[R_o] = 1

t for 0.500 = 6.66

t for 0.25 = 20 minute

d) t_1/2=1/ k[R_o]

Half time of second order reaction depends on the reactant present.Thus it changes according to reactant. No it is not consistent with time calculated in part c. The half time changes as reactant changes.

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