A chemical reaction has an activation energy of 114.882 ~kJ ~mol ⁻¹ . If the temperature is increased from 300 ~K to 600 ~K , what is the ratio of the rate constant at the higher temperature to that at the lower temperature (k₂/k₁ ) ? (Given: 2.303 R = 19.147 ~J ~K ⁻¹ ~mol ⁻¹ )

Step-by-step solution

According to the Arrhenius equation: ( k₂ k₁ ) = E_a 2.303 R ( 1 T₁ - 1 T₂ ) Given: E_a = 114.882 ~kJ ~mol ⁻¹ = 114882 ~J ~mol ⁻¹ T₁ = 300 ~K T₂ = 600 ~K 2.303 R = 19.147 ~J ~K ⁻¹ ~mol ⁻¹ Substitute the values into the equation: ( k₂ k₁ ) = 114882 19.147 ( 1 300 - 1 600 ) ( k₂ k₁ ) = 6000 ( 2 - 1 600 ) ( k₂ k₁ ) = 6000 1 600 = 10 Taking the antilogarithm (base 10) on both sides: k₂ k₁ = 10¹⁰ Answer: 10¹⁰