Question #d3bab

#A) " "k = 5.56 xx 10^(-5) "M/s"#

#B)" "[A]_0 = 5.94 xx 10^(-2) "M"#

#C)" "k = 7.58 xx 10^(-2) " s"^(-1)#

#D)" "k = 4.62 xx 10^(-2) " M"^(-1)cdot"s"^(-1)#

#A)#

In a zero-order reaction, the actual value of the concentration doesn't change what the rate is (the rate is constant, independent of concentration), so you can just take the two concentrations and times as a slope.

For some reactant #A#, in the reaction

#A -> B#,

#r(t) = -(Delta[A])/(Deltat) = -(4.00 xx 10^(-2) - 5.00 xx 10^(-2) "M")/("350 - 170 s")#

#= color(blue)(5.56 xx 10^(-5) "M/s")#

The general rate law would then be:

#r(t) = k[A]^0 = k#

since #[A]^0 = 1#, so the rate is numerically the same as the rate constant.

#B)#

The initial concentration can then be found by extrapolation back to time zero since the rate is a constant, i.e. a straight line for #[A]# vs. #t#.

If you set up a slope calculation with an initial coordinate #([A]_0,t=0)#, then you find the concentration at the y-intercept (note the slope is negative!):

#(Delta[A])/(Deltat) = -5.56 xx 10^(-5) "M/s" = ([A]_1 - [A]_0)/(t_1 - t_0)#

#= (5.00 xx 10^(-2) "M" - [A]_0)/("170 s" - "0 s")#

Therefore, solving for #[A]_0#, the initial concentration is:

#color(blue)([A]_0) = 5.00 xx 10^(-2) "M" + (5.56 xx 10^(-5) "M/s" cdot "170 s")#

#= color(blue)(5.94 xx 10^(-2) "M")#

Didn't need an integrated rate law here.

#C)#

Here, the integrated rate law has to be rederived, since aren't at time zero. Consider a first-order reaction with the rate law as follows:

#r(t) = k[A] = -(d[A])/(dt)#

Then with separation of variables:

#-kdt = 1/([A])d[A]#

Integrate from time #t_1# to #t_2# instead of #0# to #t# on the left to get:

#-int_(t_1)^(t_2) kdt = int_([A]_1)^([A]_2) 1/([A])d[A]#

#-k(t_2 - t_1) = ln[A]_2 - ln[A]_1#

And so, the more general integrated rate law is:

#color(green)(ln[A]_2 = -kDeltat + ln[A]_1)#

The rate constant for this process is then:

#color(blue)(k) = -ln(([A]_2)/([A]_1))/(Deltat) = -ln((6.40 xx 10^(-3) "M")/(9.10 xx 10^(-2) "M"))/(75.0 - "40.0 s")#

#= color(blue)(7.58 xx 10^(-2) " s"^(-1))#

#D)#

A similar rate law can be derived in general for second-order reactions. The derivation will be omitted but is a simple enough adaptation of the one with #t_1 = 0# and #t_2 = t#:

#color(green)(1/([A]_2) = kDeltat + 1/([A]_1))#

And so, its rate constant is similarly:

#color(blue)(k) = (1/([A]_2) - 1/([A]_1))/(Deltat) = (1/(2.90 xx 10^(-2) "M") - 1/("0.130 M"))/("760 s - 180 s")#

#= color(blue)(4.62 xx 10^(-2) " M"^(-1)cdot"s"^(-1))#