Question

can someone help me and explain these?

5. a) In the kinetics experiment, 12.00 mL of 0.100 M t-butyl chloride solution (in acetone) were mixed with 6.00 mL of 0.0800 M NaOH and 12.0 mL of distilled water in the presence of the bromophenol blue indicator). Calculate the composition of the reaction medium in terms of the percent water and acetone, respectively. (10 points) b) After the reaction started, the time required for the indicator to change in color from blue to yellow was measured to be 120 seconds. Calculate the value of the rate constant, k, in this medium. (10 points)

Answer 1

1) The total volume of water is 6.00 mL from the NaOH solution and 12.00 mL from the distilled water, while there are 12.00 mL of acetone from the t-butylchloride solution. The percentage of water is then:

$\frac{12.00mL+6.00mL}{12.00mL+6.00mL+12.00mL}\cdot 100=60$

60%, making the porcentage of acetone 40%.

2) The initial concentration of OH- can be calculated using the dilution formula:

$C_{f}=\frac{C_{o}\cdot V_{o}}{V_{f}}=\frac{0.0800M\cdot 6.00mL}{30.00mL}=0.016\: M$

The same for t-butylchloride:}

$C_{f}=\frac{C_{o}\cdot V_{o}}{V_{f}}=\frac{0.100M\cdot 12.00mL}{30.00mL}=0.04\: M$

I assume that the color change is observed when all of the OH- has been consumed, meaning that the average rate for the reaction is:

$rate=-\frac{\Delta [OH]}{\Delta t}=\frac{0.01600M}{120s}=1.33x10^{-4}\: \frac{M}{s}$

This is a SN1 reaction, meaning that the rate depends only on the concentration of the substrate, t-butylchloride. The rate can be expressed as:

$rate=k[t-buCl]$

Since we know the initial concentration and the average rate, we can estimate the value of k as:

$k=\frac{rate}{k[t-buCl]}=\frac{1.33x10^{-4}\frac{M}{s}}{0.04M}=0.00333\: \frac{1}{s}$