What is the rate constant at 25.0 ∘c based on the data collected for trial b?

acetone + I2 + HCl ---> iodated acetone  

Equation:

rate = k * [acetone]^x * [I2]^y * [HCl]^z  

Once we know x, y, z, we can plug in any of the trials A->D and determine k  NOTE: We can't use run E because temperature has an effect on rate. E was run at a different temperature.

The first thing to note is that do NOT have concentrations. We have volumes at a given molarity.  

Table:  

.001M I2.. ..050M HCl.. .1.0M acetone.. .water.. temp..time.. total vol  

.. mL.. ... ... .. .mL.. .. ... .. .. mL.. .. .. .. ..mL.. ..°C.. .sec.. .. .. .L  

A.. ...5.. .. .. .. .. ..10.. .. .. .. .. ..10.. .. .. .. ..25.. .. .25.. .130.. .. 0.05  

B.. ..10.. .. .. ... .. 10.. .. .. ... .. .10.. .. .. ... ..20.. . .25 249.. ..0.05  

C.. . 10.. .. .. .. .. .20.. .. .. .. 10.. .. ... ... .10... ..25.. ..128... .0.05  

D.. . 10.. ... .. ... ..10.. .. ... .. ... 20.. .. .. .. .. 10.. .. 25.. ..131.. ..0.05  

E.. ..10.. ... .. ... ..10.. ... ... ... ..10.. ... .. ... .20.. .. 42.8.. .38.. ..0.05  

We can translate that into molarity in solution using this formula. (molarity pure ingredient * mL used / 1000 / total volume in liters)  

.. .. ..I2.. HCl.. acetone.. temp.. ..rxn time  

.. .. ..M.. .. ...M.. .M.. ..°C.. .. .. sec  

A.. 0.0001.. 0.01 0.2.. .25... .130  

B.. 0.0002.. 0.01.. .. 0.2.. .. .. 25.. .. .. .249  

C.. 0.0002.. 0.02.. .. 0.2.. .. .. 25.. .. .. .128  

D.. 0.0002.. 0.01.. .. 0.4.. .. . .25.. ... ...131  

E.. 0.0002.. 0.01.. .. 0.2.. . 42.8.. .. .. 38  

From runs B and D, we can see that rate dropped by half .

As [I2] and [HCl] were held constant and [acetone] was doubled.  

This means x=-1 in this equation  

Rate = k * [acetone]⁻¹ * [I2]^y * [HCl]^z  

Rate = k * [I2]^y * [HCl]^z  

From runs A and B  

[I2] doubles  

[HCl] remains the same  

[acetone] remains the same  

rate doubles   as [I2] doubles, rate doubles  

y = 1   rate = k * [acetone]⁻¹ * [I2]¹ * [HCl]^z  

And from runs B and C, we can see that  , As [HC] doubles, (all else equal) the rate halves.

Z = -1  

Rate = k * [I2] / ([acetone] * [HCl])  

Rearranging  

k = rate * [acetone] * [HCl] / [I2]  

From any experimental run (A-D), we can calculate k.

using A to calc k... ..k = 2600 M²/sec  

using B to calc k... . k = 2490 M²/sec  

using C to calc k... . k = 2560 M²/sec  

using D to calc k... . k = 2620 M²/sec  

NOTE.. the problem statement said to use the data from run B to calc k.

Hence Final

k = 2490 M²/sec