CHM 202 Lecture Exa Fall. 2015 Answer the questions in the blue book in order. Y
ID: 561648 • Letter: C
Question
CHM 202 Lecture Exa Fall. 2015 Answer the questions in the blue book in order. You must show all working for full credit. R 0.08206 L.atm/mol.K 1. Initial rate data at 25.0 °C for the reaction NH4 (aq) + NO2 (aq)N2(g)+H20 (1) are shown below: Expt. [NH4 lo NO2lo Initial rate/Ms 0.24 0.12 0.12 0.10 0.10 0.15 7.2 x 10-6 3.6 x 10-6 5.4 x 10-6 Determine the rate law from the data and calculate the rate constant k at 25.0°C 2. At s00 °C, cyclopropane (C3Ho) rearranges to propene (CH3CH-CH2) C3H6 ()CH3CH-CH2 (g) The reaction is first order and the rate constant is 6.7 x 10-4 s-1. (a) What is the rate law for the reaction, and; (b) if the initial concentration of C3H6 is 0.100 M, what is the concentration of cyclopropane of after 20 min? 3. Butadiene (C4H6) reacts with itself at 250 C to form a dimer with the formula C3H12 2 C4H6 (g)CsH12 ( The reaction is second order in C4H6. (a) What is the rate law for the reaction, and; (b) if the rate constant is 4.0x 102 M-1.s-1, and the initial concentration of C4Hs is 0.200 M, how long will it take for the concentration of C4H6 to reach 0.04 M? a) In Q. 2 above what is the half life for the reaction and what is concentration of C3Hs after 3 half-lives 4. b) In Q. 3 above how long are the first and second half-lives?Explanation / Answer
1)NH4+ (aq) + NO2- (aq) ---->N2 (g) + 2H2O
The rate law for the reaction
rate = k [NH4+] [NO2-]
For first experiment
[NH4+] = 0.24 M
[NO2-] = 0.10 M
And initial rate = 7.2 x 10^-6 Ms-1
Rate = K [NH4+] [NO2-]
K = rate /[NH4+] [NO2-]
K = 7.2 x 10^-6 m/s /(0.24 x 0.10)
K = 3.0 x 10^-4 M/s
At 25 °C, the rate constant = 3.0 x 10^-4 M/s
2)
C3H6 --> CH3-CH =CH2
The reaction is first order in cyclopropane and has a measured rate constant of 6.7*10^-4
The initial cyclopropane concentration is 0.1 M
Cyclopropane concentration after 20 min
The rate law for the reaction
Rate = k [C3H6]
K = 6.7*10^-4
t= 20min = 1200 sec.
C3H6 initial= 0.1M
ln[C3H6o]/[ C3H6t] = - kt
ln [C3H6t] = (6.7*10^-4) x (1200) + ln [0.1]
ln [C3H6t] = 0.804 + (-2.3)
ln [C3H6t] = - 1.496
[C3H6t] = 10^-(1.496)
[C3H6t] = 0.0319 M
Cyclopropane concentration after 20 min is 0.032 M