Question

At what temperature k1 = K2​

Answer 1

Answer:

Since you need to know at least  one temperature at the following reaction takes place,

I have derived the relation between them here:

$\\\tt \dfrac{ {2} }{T_{2} } - \dfrac{ 1 }{T_{1}} = \dfrac{R}{2.303 \times 10^{4}}$

Explanation:

Using Arrhenius equation:

$\\\tt k=Ae^{{\dfrac {-E_{a}}{RT}}}$

$\\\tt log k = log A - \dfrac{2.303 \ E_{a}}{RT}$

Using values for $\\\tt k_{1}$

$\\\tt k_{1} = log (10^{14}) - \dfrac{2.303 \times 10^{4} }{RT_{1}}$

and

$\\\tt k_{2} = log (10^{15}) - \dfrac{2.303 \times 20^{4} }{RT_{2}}$

For them to be equal:

$\\\tt log (10^{14}) - \dfrac{2.303 \times 10^{4} }{RT_{1}} = log (10^{15}) - \dfrac{2.303 \times 20^{4} }{RT_{2}}$

$\\\tt \implies 14 - \dfrac{2.303 \times 10^{4} }{RT_{1}} = 15 - \dfrac{2.303 \times 20^{4} }{RT_{2}}$

$\\\tt -1= \dfrac{2.303 \times 10^{4} }{RT_{1}} - \dfrac{2.303 \times 20^{4} }{RT_{2}}\\-1= \dfrac{2.303 }{R} \times [ \dfrac{ 10^{4} }{T_{1} } - \dfrac{ 20^{4} }{T_{2}} ]$

$\\\tt -1= \dfrac{2.303 }{R} \times [ \dfrac{ 10^{4} }{T_{1} } - \dfrac{ 20^{4} }{T_{2}} ] \\\\\tt \dfrac{ {2} }{T_{2} } - \dfrac{ 1 }{T_{1}} = \dfrac{R}{2.303 \times 10^{4}}$