The reaction A-->B+C is second order in A .When the initial concentration of A=.100 M the reaction is 20%?

Question

complete in 40 minutes. Calculate the value of the rate constant (in L/min mol)

Answer 1

dA/dt = -kA^2

dA/A^2 = -kdt
1/A = -kt + C
1/.100 = 0 + C
C = 10

A = 1/ ( -kt + 10)

80% of A remains after 40 minutes
.80 * .100M = .080M

0.080 = 1/ ( -k*40 + 10)
( -k*40 + 10) = 12.5
k = -0.0625

Answer 2

You might try asking this question in the science category. I'm a little rusty on my chemistry and organic chemistry, but I know that the answer lies in the fact that .100 M is used and that 20% of the chemicals react at that concentration. I hope this helps :-)