Question

At 25∘C, 0.11 mol of N2O4 reacts to form 0.10 mol of N2O4 and 0.02 mol of NO2. At 90∘C, 0.11 mol of N2O4 forms 0.050 mol of N2O4 and 0.12 mol of NO2. From these data, it can be concluded that
A. N2O4 molecules react by a second order rate law.
B. N2O4 molecules react by a first order rate law.
C. N2O4 molecules react faster at 25∘C than at 90∘C.
D. the equilibrium constant for the reaction above increases with an increase in temperature.

Answer 1

D. Determining first or second order isn't done with different temperatures so neither A nor B are right.
C isn't right because increaing T increases the rate of reaction (usually).
So D must the the correct answer. Calculate Keq and see it K increases with T. .........N2O4 --> 2NO2
@25C I..........0.11........0
............E.........0.10.....0.02
K@25 C = (NO2)^2/(N2O4) = (0.02)^2/0.10 = ?
@90C I..............0.11........0
............E............0.05........0.12
K@90C = (0.12)^2/0.05 = ?
Compare the two K values at the two temperatures.

Answer 2

Thank you SOO much!