Part A The rate constant for a certain reaction is k = #7.00xx10^-3# #s^-1#. If the initial reactant concentration was 0.600 M, what will the concentration be after 19.0 minutes?

1 Answer
Jul 15, 2016

Answer:

#sf(2.05xx10^(-4)color(white)(x)"mol/l")#

Explanation:

The units for #sf(k)# tells me that the reaction is 1st order.

So we get:

#sf(Ararr"products")#

#:.##sf(-(d[A])/dt=k[A]^1)#

Integrating gives:

#sf(ln[A]_t=ln[A_0]-kt)#

Putting in the numbers:

#sf(ln[A]_t=ln[0.6]-(7xx10^(-3)xx19.0xx60))#

#sf(ln[A_t]=-0.511-7.98=-8.491)#

From which:

#sf([A]_t=2.05xx10^(-4)color(white)(x)"mol/l")#

This is 1st order exponential decay. The concentration / time graph looks like this:

www.cliffsnotes.com

You will notice that the time taken for #sf(A)# to fall by 1/2 its initial value is constant.

This is a feature of 1st order processes.