The second order reaction 2A 2 B 3 -----> A 4 B 6 has a rate constant, k, of 7.3
ID: 891637 • Letter: T
Question
The second order reaction 2A2B3 -----> A4B6 has a rate constant, k, of 7.35 x 10-3M-1S-1 at a certain temperature. If [A2B3]0 = 1.32M, how much time (s) does it take to reduce the concentration to 27% of the original? (please explain)
The decomposition of hydrogen peroxide H2O2(l) ---> H2(g) + O2(g) has a rate constant of 1.06 x 10-3min-1. how long (min) will it take for the concentration of H2O2 to decompose by 60%? (please explain)
the decomposition of M3X2(aq) <--> 3M(s) + X2(g) has an activation energy of 15- kJ/mol and (delta)Hrxn = -72kJ/mol, under specified conditions. what is the activation energy (kJ/mol) for the reverse reaction?
the equilibriium constant, Kp has a value of 6.5x10-4 at 308 K for the reaction of nitrogen monoxide with chlorine. 2NO(g)+Cl2(g) <---> 2NOCl(g)... at 308K what is the value of Kc? (please explain)
Explanation / Answer
1) The second order reaction 2A2B3 -----> A4B6 has a rate constant, k, of 7.35 x 10-3M-1S-1 at a certain temperature. If [A2B3]0 = 1.32M, how much time (s) does it take to reduce the concentration to 27% of the original? (please explain)
Solution :-
Formula of the second order kinetics is as follows
1/[A]t = K*t + 1/[A]o
Concentration reduced to 27% of the original therefore if we assume 100 % at initial then final is 100 -27 = 73 %
1.32 M * 0.73 = 0.9636 M
Now lets put the values in the formula
1/[A]t = K*t + 1/[A]o
1/[0.9636] = 7.35*10^-3 M-1 s-1 * t + 1/[1.32]
1.04 = 7.35*10^-3 M-1 s-1 * t + 0.7575
1.04-0.7575 = 7.35*10^-3 M-1 s-1 * t
0.2802 /7.35*10^-3 M-1 s-1 = t
38.1 sec =t
Time needed = 38.1 s
2) The decomposition of hydrogen peroxide H2O2(l) ---> H2(g) + O2(g) has a rate constant of 1.06 x 10-3min-1. how long (min) will it take for the concentration of H2O2 to decompose by 60%? (please explain)
Solution :- Reaction is first order
Using the first order rate equation we can calculate the time
ln([A]t/[A]o) = - k*t
concentration decreasing by the 60 % means remaining is 100 – 60 = 40 %
so lets put the values in the formula
ln[40/100]= - 1.06*10^3 min-1 * t
-0.9163 = -1.06*10^-3 min-1 * t
-0.9163 / -1.06*10^-3 min-1 = t
864.4 min = t
Therefore time needed is 864.4 min
3) the decomposition of M3X2(aq) <--> 3M(s) + X2(g) has an activation energy of 15- kJ/mol and (delta)Hrxn = -72kJ/mol, under specified conditions. what is the activation energy (kJ/mol) for the reverse reaction?
Solution :- When the reaction is reversed reaction is as follows
Ea reversed rrxn = 15 kJ + 72 kJ = 87 kJ per mol
4) the equilibriium constant, Kp has a value of 6.5x10-4 at 308 K for the reaction of nitrogen monoxide with chlorine. 2NO(g)+Cl2(g) <---> 2NOCl(g)... at 308K what is the value of Kc?
Solution :-
Formula to calculate the Kc from the kp is as follows
Kp = Kc*(RT)^delta n
Delta n = 2- (2+1) =-1
Kc = Kp/RT^delta n
Kc = 6.5*10^-4 / (0.08206 L atm per mol K * 308 K)^-1
Kc= 1.64*10^-2
Therefore the KCc of reaction = 1.64*10^-2