A second-order reaction starts with an initial concentration of 0.100 mol/L of the reactant.If the rate constant is 1.4 $\times$ 10^-2 L/mol·s,what is the time required to decrease the initial concentration to 0.050 mol/L?
2A $\to$ B $~~~~~~~~$ rate = k[A]²

Question

A second-order reaction starts with an initial concentration of 0.100 mol/L of the reactant.If the rate constant is 1.4

\times

10^-2 L/mol·s,what is the time required to decrease the initial concentration to 0.050 mol/L?
2A

\to

B

~~~~~~~~

rate = k[A]²

Quizplus · Accepted Answer

For a second-order reaction, the integrated rate law is 1/[A] = kt + 1/[A]₀. Plugging in the values: 1/0.050 = (1.4 × 10⁻²)t + 1/0.100. Solving for t gives t = 710 s.

A Second-Order Reaction Starts with an Initial Concentration of 0 ×\times×

A Second-Order Reaction Starts with an Initial Concentration of 0 $\times$