A reaction takes place at $300 K$ . When the catalyst is added rate of reaction increases. Calculate the change in temperature $($ in $^{o} C)$ which would have the same effect on the rate of reaction as that of catalyst. (Assume catalyst change the activation energy by $20 %$ )

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 75
$ℓ n \frac{K_{2}}{K_{1}} = \frac{E_{a}}{R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$ ............ (i)

$ℓ n \frac{K_{C}}{K_{1}} = \frac{E_{a} - E_{a c}}{R T_{1}}$

But, for effect of catalyst new $K_{C} = K_{2}$

$ℓ n \frac{K_{2}}{K_{1}} = \frac{E_{a} - E_{a c}}{R T_{1}}$ ...........(ii)

from (i) and (ii), we get,

$\frac{E_{a} - E_{a c}}{R T_{1}} = \frac{E_{a}}{R} . [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

$\frac{1}{T_{1}} - \frac{1}{T_{2}} = \frac{E_{a} - E_{a c}}{E_{a} . T_{1}}$

$\frac{1}{300} - \frac{1}{T_{2}} = \frac{100 - 80}{100 \times 300} = \frac{20}{100 \times 300} = \frac{1}{1500}$

$\frac{1}{T_{2}} = \frac{1}{300} - \frac{1}{1500} = \frac{5 - 1}{1500} = \frac{4}{1500}$

$T_{2} = 375$

Change in temp $= 375 - 300 = 75$

Hence, the correct option is $A$

Suggest Corrections

Similar questions

(i n^{\circ} C)

20 %

x

\frac{x + 6}{9}

27^{o} C

1000

500 K

400 K

20 kJ/mol

20 k J m o l^{- 1}

Join BYJU'S Learning Program