Question

C4H8(g)⟶2C2H4(g) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ/mol.262 kJ/mol. At 600.0 K,600.0 K, the rate constant, ????,k, is 6.1×10−8 s−1.6.1×10−8 s−1. What is the value of the rate constant at 725.0 K?

Answer 1

Answer:

Rate constant at 725 K is $5.2\times 10^{-4}s^{-1}$

Explanation:

According to Arrhenius equation for a reaction-

$ln(\frac{k_{2}}{k_{1}})=\frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})$

where $k_{2}$ and $k_{1}$ are rate constants of reaction at $T_{2}$ and $T_{1}$ temperatures (in kelvin) respectively.

$E_{a}$ is activation energy of reaction.

Here $T_{1}$= 600 K , $k_{1}$= $6.1\times 10^{-8}s^{-1}$

$T_{2}$= 725 K, $E_{a}$= 262 kJ/mol and R = 8.314 J/(mol.K)

So plugin all the values in the above equation-

$ln(\frac{k_{2}}{6.1\times 10^{-8}s^{-1}})=\frac{262\times 10^{3}J/mol}{8.314J/(mol.K)}\times (\frac{1}{600K}-\frac{1}{725K})$

So, $k_{2}$ = $5.2\times 10^{-4}s^{-1}$

Hence rate constant at 725 K is $5.2\times 10^{-4}s^{-1}$