Question

A reactant. Q. decomposes at a second order. The slope of the graph 1/[Q] (1/M) vs time (s) is -0.04556. If the initial

concentration of Q for the reaction is 0.50 M, what is the concentration in M. of Q after 10.0 minutes?

Answer 1

0.034 M

Step-by-step explanation:

1/[A] = kt + 1/[A]o

[A] = ?

k= 0.04556

t= 10.0 minutes or 600 seconds

[A]o = 0.50 M

1/[A] = (0.04556 × 600) + 1/0.50

[A] = 0.034 M