**a). Rate expression**

The total volume in each experiment is the same: 50 mL.

Hence, the concentrations of #"KI"# and #"Fe"^"3+"# are directly proportional to their volumes.

We can't get the order with respect to #"KI"# because its concentration is always the same.

However, we **can** get the order with respect to #"Fe"^"3+"#.

We are given

#y ∝ ["Fe"^"3+"]^x# and

#y ∝ 1/t["Fe"^"3+"]^x#

∴ #y ∝ 1/t# or #y = k/t#

From Experiments 1 and 2, it appears that doubling the concentration halves the time (doubles the rate).

Thus the reaction appears to be first order in #["Fe"^"3+"]#.

Hence, the rate expression is #r = k["Fe"^"3+"]^1# or #r = k["Fe"^"3+"]#.

**(b) Plot of concentration vs #1/t#**

If a plot of rate (#1/t#) vs concentration (#x#) is a straight line, the reaction is first order in #["Fe"^"3+"]#.

We have the following data:

#bb("Expt."color(white)(m)x//"cm"^3color(white)(m) t color(white)(mml)1//t)#

#color(white)(m)1color(white)(mmmml)5color(white)(mm)92color(white)(mm)0.0109#

#color(white)(m)2color(white)(mmmll)10color(white)(mm)45.4color(white)(ml)0.0220#

#color(white)(m)3color(white)(mmmll)15color(white)(mmll)?color(white)(mmm)?#

#color(white)(m)4color(white)(mmmll) 20color(white)(mm)23color(white)(mml)0.0435#

#color(white)(m)5color(white)(mmmll)25color(white)(mm)18.1color(white)(ml)0.0552#

We plot #1/t# vs #x# and get a graph that looks like this:

You are given that the equation for the line is #y = kx#, so #k# is the slope of the graph.

The points at #"5 cm"^3# and #"25 cm"^3# appear to be on the line, so we will use them to calculate the slope.

#"Slope" = (Δy)/(Δx) = (y_2-y_1)/(x_2-x_1) = (0.0552 - 0.0109)/("25 cm"^3 - "5 cm"^3) = 0.0443/("20 cm"^3) = 2.22 × 10^"-3"color(white)(l) "cm"^"-3"#

Thus, the rate expression from the graph is

#r = 2.22 × 10^"-3"color(white)(l) "cm"^"-3" × ["Fe"^"3+"]#.

Since #r# is directly proportional to #["Fe"^"3+"]#, the reaction is first order.

The calculated value of #1/t# for #y = "15 cm"^3# is

#1/t = 2.22 × 10^"-3" color(red)(cancel(color(black)("cm"^"-3"))) × 15 color(red)(cancel(color(black)("cm"^3))) = 0.0333#

(I have shown it as a brown point on the line.)

∴ #t = 1/0.0333 = 30.0#