Consider the reaction 2o3(g)  ⟶3o2(g);   use the following data to calculate the activation energy for the reaction.          t(°c)                                    rate constant (m-1s-1)        527                                        3.58  x  105      727                                        5.90 x 10644.6 kj/mol920 j/mol440 j/mol93.2 kj/mol

Eₐ = 44.6 kJ·mol⁻¹

Use the Arrhenius equation

ln(k₂/k₁) = (Eₐ/R)(1/T₁ -1/T₂)

ln(5.90× 10⁶/3.58 × 10⁵) = (Eₐ/8.314 J·K⁻¹mol⁻¹) (1/527 K - 1/727 K)  

ln16.48 = (Eₐ/8.314J·mol⁻¹) × 5.220 × 10⁻⁴

2.802 = Eₐ ×6.729 × 10⁻⁵ mol·J⁻¹

Eₐ = 2.802/(6.729 × 10⁻⁵ mol·J⁻¹) = 4.46 × 10⁴ J·mol⁻¹

Eₐ = 44.6 kJ·mol⁻¹

oh it is true if thats what you are asking

8 m/s because for him to get the same amount of energy as ray is putting out, bobby, who weighs half as much as ray, needs to run twice as fast.