The activation energy for the isomerization ol cyclopropane to propene is 274 kJ/mol. By what factor does the rate of this reaction increase as the temperature rises from 255 to 291 oC? Hint: The factor is the ratio of the rates. Since rate is directly proportional to rate constant, the factor is also ratio of rate constants.

53.73 factor

Explanation:

Given that:

The activation energy () =  274 kJ/mol.

T₁ = 255°C  = (255+273) = 528 K

T₂ = 291° C = (291 + 273) = 564 K

Rate Constant (R) = 8.314 × 10⁻³ kJ k⁻¹ mol⁻¹

We are tasked to determine the factor at which the rate of the reaction increase as the temperature rises from 255°C to 291° C; In order to do that;

We can apply the use of Arrhenius Equation, which is given by:

㏑ =

㏑ =

㏑ =

㏑ = 3.984

 =

= 53.73

K₂ = 53.73 K₁

∴ As the temperature rises from 255°C to 291° C, the rate of reaction increases by 53.73 factor

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.