Question

The rate constant for a second-order reaction is 0.13 m-1 sec-1. if the initial concentration of reactant is 0.26 mol/l, it takes __________ sec for the concentration to decrease to 0.13 mol/l.

Answer 1

The time it takes for the concentration to decrease to 0.13 mol/L is approximately 6.13 seconds.

Step-by-step explanation:

The rate constant for a second-order reaction is 0.13 m-1 sec-1. If the initial concentration of reactant is 0.26 mol/L, we can use the integrated rate law for a second-order reaction to determine the time it takes for the concentration to decrease to 0.13 mol/L.

The integrated rate law for a second-order reaction is:
[A] = [A0] / (1 + k[A0]t)

Substituting the given values:
0.13 mol/L = 0.26 mol/L / (1 + 0.13 mol-1 sec-1 × 0.26 mol/L × t)

Solving for t:
0.13 = 0.26 / (1 + 0.0338t)

0.0338t = 0.26 - 0.26 × 0.13
0.0338t = 0.26(1 - 0.13)
t = 0.26(1 - 0.13) / 0.0338
t ≈ 6.13 seconds

Answer 2

The number of seconds for the concentration to decrease to 0.13 mol for a second order reaction is calculated using the below formula

1/At =-Kt +1/Ao where At is the final concentration, Ao is the initial concentration. t = time in second while k = constant

1/0.13 =-0.13 (t) + 1/0.26
like terms together

1/0.13-1/0.26 =-0.13 t
divide both side by 0.13

T =29.6 seconds