For a first-order reaction, graphs of ( P₀ P ) versus time t are plotted at two different temperatures, yielding two straight lines passing through the origin. At T₁ = 300 K, the slope of the line is 2 10⁻³ s ⁻¹ . At T₂ = 400 K, the slope of the line is 5.436 10⁻³ s ⁻¹ . The value of E_a R for this reaction is ________ K. (Given: e 2.718 , where E_a is the activation energy and R is the universal gas constant)

Step-by-step solution

For a first order reaction, the integrated rate law can be written as: ( P₀ P ) = kt The slope of the graph of ( P₀ P ) versus t is equal to the rate constant k . At T₁ = 300 K, k₁ = 2 10⁻³ s ⁻¹ At T₂ = 400 K, k₂ = 5.436 10⁻³ s ⁻¹ Taking the ratio of the rate constants: k₂ k₁ = 5.436 10⁻³ 2 10⁻³ = 2.718 e According to the Arrhenius equation: ( k₂ k₁ ) = E_a R ( 1 T₁ - 1 T₂ ) (e) = E_a R ( 1 300 - 1 400 ) 1 = E_a R ( 4 - 3 1200 ) 1 = E_a R 1 1200 E_a R = 1200 K. Answer: 1200