A. The concentration ratio between trials 2 and 1 is: 0.3 / 0.1 = 3.
B. The ratio between reaction rates is: 0.054 / 0.006 = 9.
C. We solve the equation: (concentration ratio)^x = (reaction rate ratio):
, which gives x = 2, and the exponent in the rate law is 2.
D. The rate law will have the equation: r = k[NO2]^2
E. To solve for k, we can substitute the values for [NO2] = 0.1, rate = 0.006. This gives:
, which yields k = 0.6 L/mol-s.
F. Since the exponent is 2, this is a second-order reaction.
B. The ratio between reaction rates is: 0.054 / 0.006 = 9.
C. We solve the equation: (concentration ratio)^x = (reaction rate ratio):
D. The rate law will have the equation: r = k[NO2]^2
E. To solve for k, we can substitute the values for [NO2] = 0.1, rate = 0.006. This gives:
F. Since the exponent is 2, this is a second-order reaction.