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.