Question

For reactions $P\to Q$ and $X\to Y$ Arrhenius constants are ${10}^6$ and ${10}^8$ respectively. If $E_{P\to Q}=1500\text{cal/mole}$ and $E_{x\to y}=2000\text{cal/mole}$ then find the temperature at which their rate constants are same. (Given $R=2\text{cal/mole.K}$)

Answer 1

Arrhenius equation:
$k=A{e}^{-E_a/RT}$
Where,
$k=\text{Rate constant}$
$A=\text{Arrhenius constant}$
$R=\text{Gas constant}$
$T=\text{Temperature}$
$P⟶Q...\left(i\right)$
$A_1={10}^6,E_{a1}=1500{\text{cal mol}}^{-1}$
$k_1={10}^6{e}^{-1500/RT}$
$X⟶Y...\left(ii\right)$
$A_2={10}^8,E_{a2}=2000{\text{cal mol}}^{-1}$
$k_2={10}^8{e}^{-2000/RT}$
According to question:
$k_1=k_2$
$\Rightarrow {10}^6{e}^{-1500/RT}={10}^8{e}^{-2000/RT}$
$\Rightarrow {10}^{-2}=e^{(1500-2000)\left[\frac{1}{2T}\right]}$
$\Rightarrow {10}^{-2}=e^{(-250)\left[\frac{1}{T}\right]}$
On taking algorithm on both sides:
$\Rightarrow T=\frac{250}{4.606}\text{K}$