rate = (130)(0.2)^2 (0.03) = 0.156
Again, even if we had not known what the rate constant is, we would not need to
calculate it explicitly for this problem. If [SO 3 ] were raised to 0.2 in trial 4, it would have
been doubled. Therefore the rate would be quadrupled, and would equal 4 × 0.039 =
0.156.6 . B
The activation energy is the minimum amount of energy needed for a reaction to proceed. The
activation energy for the reverse reaction is the change in potential energy between the
products and the transition state indicated by arrow B. A is the overall (thermodynamic)
energy change of the reaction. C is the activation energy for the forward, not the reverse,
reaction.
7 . The best way to visualize the solution to this set of problems is to draw a diagram.
ΔE = Activation Energy (forward) – Activation Energy (reverse)= 78 kJ – 300 kJ = −222 kJ.
And sinceΔE = Energy (products) − Energy (reactants)−222 kJ = 25 kJ − X kJ,X = 247 kJ = energy of reactants.