Consider the data below for the reaction X ? products. (Hint: You will have to d
ID: 600270 • Letter: C
Question
Consider the data below for the reaction X ? products. (Hint: You will have to determine the order of the reaction before you can answer parts (b)-(d) of this question.)=Time (in seconds)
=[X] (in mol/L)
Time=0
[X]=1.7500
Time=12.6
[X]=1.1708
Time=25.3
[X]=0.7808
Time=37.9
[X]=0.5224
Time=50.5
[X]=0.3495
(a) What is the average rate of consumption of X between 12.6 s and 50.5 s? (Enter your answer accurate three significant figures. use concentrations in mol/L and time in seconds.) ____________
(b) What is the order of the reaction? (Enter an integer, or a fraction, e.g. 1/2, 3/2, etc.) ____________
(c) What is the value of the rate constant, k?answer accurate to three significant figures. Use concentrations in mol/L and time in seconds.) ____________
(d) What is the instantaneous rate of consumption of X at 50.5 s? answer accurate to three significant figures.Use concentrations in mol/L and time in seconds.) ____________
Explanation / Answer
take help from this solved example and visit this link http://www.google.com.pk/url?sa=t&rct=j&q=What+is+the+value+of+the+rate+constant%2C+k%3Fanswer+accurate+to+three+significant+figures.+Use+concentrations+in+mol%2FL+and+time+in+seconds.)&source=web&cd=3&cad=rja&ved=0CC8QFjAC&url=http%3A%2F%2Fteachers.sduhsd.net%2Fkshakeri%2FAP_Chemistry_Pages%2FHandouts%2FChapter_14%2Fchapter14.pdf&ei=dNpvUMGkMJDKswaktICoDg&usg=AFQjCNHos-ZyG-5_BccsaW4y_oaHG5i7tA 14.1 Rate of formation of NO2F = ?[NO2F]/?t. Rate of reaction of NO2 = -?[NO2]/?t. Divide each rate by the coefficient of the corresponding substance in the equation: 1/22[NOF]t?? = -1/22[NO]t??; or 2[NOF]t?? = - 2[NO]t?? 14.2 Rate = - -[I]t?? = - [0.00101 M - 0.00169 M]8.00 s - 2.00 s? = 1.13 x 10-4 = 1.1 x 10-4 M/s 14.3 The order with respect to CO is zero, and with respect to NO2 is two. The overall order is two, the sum of the exponents in the rate law. 14.4 By comparing experiments 1 and 2, you see the rate is quadrupled when the [NO2] is doubled. Thus, the reaction is second order in NO2, and the rate law is Rate = k[NO2]2 (continued) RATES OF REACTION ?? 467 The rate constant may be found by substituting experimental values into the rate-law expression. Based on values from Experiment 1, k = 22rate[NO] = 27.1 x 10 mol/(Ls)(0.010 mol/L)• = 0.710 = 0.71 L/(mol•s)