Question

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.

Answer 1

Answer:

53.73 factor

Explanation:

Given that:

The activation energy ($E_a$) =  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:

㏑ $(\frac{K_2}{K_1})$ = $\frac{E_a}{R} [\frac{1}{T_1}-\frac{1}{T_2}]$

㏑ $(\frac{K_2}{K_1})$ = $\frac{273kJ/mol}{8.314*10^{-3}kJK^{-1}} [\frac{1}{528}-\frac{1}{564}]$

㏑ $(\frac{K_2}{K_1})$ = $\frac{273kJ/mol}{8.314*10^{-3}kJK^{-1}} [\frac{36k}{(528K)(564K)}]$

㏑ $(\frac{K_2}{K_1})$ = 3.984

$(\frac{K_2}{K_1})$  = $e^{3.984$

$(\frac{K_2}{K_1})$ = 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