Question

The following reaction was monitored as a function of time:
AB→A+B
A plot of 1/[AB] versus time yields a straight line with slope 5.7×10⁻² (M⋅s)−1 .

Part A

What is the value of the rate constant (k) for this reaction at this temperature?

Express your answer using two significant figures.

Part B

Write the rate law for the reaction.

Part C

What is the half-life when the initial concentration is 0.59 M ?

Express your answer using two significant figures.

Part D

If the initial concentration of AB is 0.260 M , and the reaction mixture initially contains no products, what are the concentrations of A and Bafter 75  s ?

Express your answers numerically using two significant figures, separated by a comma.

Answer 1

for any order -d[AB]/dt= K[AB]ⁿ, n= order of reaction

for second order reaction, -d[AB]/dt= K[AB]², when the equation is integrated

1/[AB] =1/[AB]0+ Kt   (1) where [AB]0 =initial concentration of AB, [AB]= concentration of AB at any time t, K = rate constant

so a plot 1/[AB] vs time gives slope. Hence K= 5.7*10^-2/M.sec ( the units of K itself tells that the reaction is seond order reaction)

hence the rate equation is -dAB/dt= 5.7*10^-2[AB]²,

half life (t1/2) is defined as the time required for the concentration to drop to 50% of initial value. Accordingly

Eq.1 becomes 1/[AB]o/2= 1/[AB]o + K*t1/2

1/[AB]0= K*t1/2

t1/2= (1/0.59M*5.7*10^-2/M.sec)=29.73 sec

when [AB]0= 0.26M, t= 75 sec, from Eq.1

1/[AB]= 1/0.26+5.7*10^-2*75

[AB]= 0.123 M