itales, Uhder, and activation energy of a chemical reaction, Part A A. Obiective

Question

itales, Uhder, and activation energy of a chemical reaction, Part A A. Obiective: What is the objective of Part A of Experiment 11? b hire effect of concemivation on rate th remotion. deter What is the complete balanced equation for the reaction studied in Exp 11? B. Data: Complete the following Table 11.2 based on your experimental results. Mixture Temp (PC) Time (s) Relative Rate, Initia ConcentratonsinmbetureM 1000/time 22.4 C 2s 20 s 50 22.0 C. Calculations: 1. Show how you calculated allthe concentrations for Reaction Mixture 2. Bras 0loM 20 mL/so mL .020 M 0080 M 00a on 2. Show how you calculated the order of the reaction with respect to Broa, ie, the value of b.

Explanation / Answer

C. Calculations:

2. The order of reaction with respect to [BrO3-] can be calculated as follows:

Comparing mixture 1 and 2 , we can see that the concentrations of [I-] and [H+] are constant and concentration of [BrO3-] is changed.

Rate = k[I-]x[BrO3-]y[H+]z

So, rate 3/rate 1 = k(0.0020M)x(0.0160M)y(0.020M)z / k(0.0020)x(0.0080)y(0.020)z

25M/s / 12.5M/s = (0.0160M/0.0080M)y

2 = 2y

therefore, y=1

Similarly we can determine the order of reaction with respect to [I-] and [H+]

rate2/rate 1 = 25.6M/s / 12.5M/s = (0.0040M/0.0020)x

2 = 2x

so, x =1

For [H+] , rate4/rate1 = 50M/s / 12.5M/s = (0.040M/0.020)z

4=2z

So, z = 2

3. Now that we know x, y and z, we can use rate law to determine the rate constant.

Mixture 1 : rate = k[0.0020M]1[0.0080M]1[0.020M]2

12.5M/s = k * 6.4*10-9 M4

k = 2.0 * 109 M-3s-1

Similarly calculate for mixture 2, 3 and 4 and then take the average.

mixture 2 : k =2.0*109 M-3s-1

mixture 3 : k = 2.0*109 M-3s-1

mixture 4: k = 2.0*109 M-3s-1

kaverage = 2.0*109 M-3s-1

E. 1. a) Now using kaverage value, calculate the predicted relative rate for mixture 5

relative rate = k * [0.0016M]1[0.0040]1[0.030]2

= 2.0*109 M-3s-1*5.76*10-9 M4

   = 11.5 Ms-1

b) now calculate the predicted time

relative rate = 1000/t = 11.5 Ms-1

t = 87 s

c) predicted time = 87 s

observed time = 82 s

% difference = (87-82)s/(82+87/2)s * 100 = 5/84.5*100 =5.9%

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