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81 What data should be plotted to show that experimental concentration data fits a second- order reaction? ANSWER: A) 1/(reactant) vs. time B) [reactant) vs. time C) In[reactant) vs. time C) In(k) vs. 1/T E) In(k) vs. Ea 9) Given the following rate law, how does the rate of reaction change if the concentration of Y is doubled? Rate = k [X]2[M3 ANSWER: 10) A reaction is followed and found to have a rate constant of 3.36 x 104 M-13-1 at 344 Kanda rate constant of 7.69 M-lg-1 at 219 K. Determine the activation energy for this reaction. ANSWER:

Answer 1

For a 2nd order she reaction, equation relation concentration to time is given by, [A reactant concentration 1 -1 = kt! TAJ TAJ at time t If we plot a graph of t us Treactant] time, we get a straight line which shows son is of and order. Opton A is correct. a. Rate = K[x]²[y] If we double concentration of keeping concentration of x same. New Rate = K[x]²[2y3 = 8 k{x}}x} New Rate becomes 8 times of old rate 10. at T = 344k RT Given, K = 3.36X10M's & K= 7.69 M5 at T2 = 219k Using Arsohenius law, K= A exp(-Ea ) for two different temperature, we get en (2) Pan (+7) ln / 7.69, liste On solving, Actiration energy = Ea = 45.654 kJ mo