From K=Ae^ Ea/RT , We know that K_1K_2=T_1T_2 #160; #160; #160; #160;( K = number of molecules, T = Temperature) ⇒5025=300T_2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Activation Energy

At 300 K, 5...

Question

At $300 K, 50$ of molecule collide with energy grater than or equal to $E_{a}$ . At what temperature, $25$ molecule will have energy greater than or equal to $E_{a}$ .

Open in App

Solution

From $K = A e^{- E a / R T}$ , We know that $\frac{K_{1}}{K_{2}} = \frac{T_{1}}{T_{2}}$ ( $K$ = number of molecules, $T$ = Temperature)
$\Rightarrow \frac{50}{25} = \frac{300}{T_{2}}$
$\Rightarrow T_{2} = 150 K$

Suggest Corrections

Similar questions

E_{a}

E_{a}

E_{a}

H_{2} O_{2} \to H_{2} O + \frac{1}{2} O_{2}

18 K c a l / m o l

300 K

(- 13.02) = 9.36 \times 10^{- 14}

In the equation k = Z_AB e^-E_a^/RT, e^-E_a^/RT represents

Join BYJU'S Learning Program