The activation energy for the gas phase decomposition of 2-bromopropane is 212 kJ.

CH3CHBrCH3CH3CH=CH2 + HBr

The rate constant at 683 K is 6.06*10-4 / s. The rate constant will be 5.06*10-3 / s at

The question is essentially asking for the temperature at which the rate constant will be equal to 5.06 x 10⁻³. We will be using the Arrhenius equation to solve this problem. Which is shown below:

k = Ae^ (-Ea/RT)

We first must use the values provided to solve for the value of A. Once we know A, we can solve for the temperature at a given rate constant.

A = k / (e^ (-Ea/RT))

A = (6.06 x 10⁻⁴) / (e^ (212000 / (8.314 x 683)))

A = 9.92 x 10⁻¹²

Now that we have a value for A, we can solve for the temperature that gives the provided rate constant. First we can rearrange the equation to solve for T:

k = Ae^ (-Ea/RT)

lnk = lnA - Ea/RT

lnk - lnA = - Ea/RT

T = - Ea / (R (lnk - lnA))

T = - 212 / (0.008314) (ln (5.06 x 10⁻³) - ln (9.92 x 10⁻¹²))

T = 724 K

The rate will be equal to 5.06 x 10⁻³ at a temperature of 724 K.