Question

The Arrhenius equation shows the relationship between the rate constant k and the temperature T in kelvins and is typically written as k=Ae−Ea/RT where R is the gas constant (8.314 J/mol⋅K), A is a constant called the frequency factor, and Ea is the activation energy for the reaction. However, a more practical form of this equation is lnk2k1=EaR(1T1−1T2) which is mathematically equivalent to lnk1k2=EaR(1T2−1T1) where k1 and k2 are the rate constants for a single reaction at two different absolute temperatures (T1 and T2). Part A The activation energy of a certain reaction is 45.6 kJ/mol . At 30 ∘C , the rate constant is 0.0160s−1 . At what temperature in degrees Celsius would this reaction go twice as fast

Answer 1

Answer:

T = 42.08  °C

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Wherem  

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314 J / K mol

Thus, given that, $E_a$ = 45.6 kJ/mol = 45600 J/mol (As 1 kJ = 1000 J)

$k_2=2\times k_1$

$k_1=0.0160s^{-1}$

$T_1=30\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (30 + 273.15) K = 303.15 K  

$T_1=303.15\ K$

So,

$\ln \dfrac{k_{1}}{2\times k_{1}} =-\dfrac{45600}{8.314} \left (\dfrac{1}{303.15}-\dfrac{1}{T_2} \right )$

$\ln \dfrac{1}{2} =-\dfrac{45600}{8.314} \left (\dfrac{1}{303.15}-\dfrac{1}{T_2} \right )$

$8.314\ln \left(2\right)=-45600\left(\frac{1}{303.15}-\frac{1}{T_2}\right)$

$8.314\ln \left(2\right)=-150.42058+\frac{45600}{T_2}$

$144.65775 =\frac{45600}{T_2}$

$T_2=\frac{45600}{144.65775}$

$T_2=315.23\ K$

Conversion to °C as:

T(K) = T( °C) + 273.15  

So,  

315.23 = T( °C) + 273.15

T = 42.08  °C