How fast is the reaction at 25, ^circ C as co | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Using the Arrhenius equation: $\log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303 R} \left( \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right)$Given: $E_{a} = 6

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(C) Using the Arrhenius equation: $\log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303 R} \left( \frac{T_{2} - T_{1}}{T_{1} T_{2}} ight)$Given: $E_{a} = 65000\, J/mol$,$T_{1} = 273\, K$,$T_{2} = 298\, K$,$R = 8.314\, J/mol\cdot K$.Substituting the values:$\log \frac{K_{2}}{K_{1}} = \frac{65000}{2.303 	imes 8.314} \left( \frac{298 - 273}{273 	imes 298} ight)$$\log \frac{K_{2}}{K_{1}} = \frac{65000}{19.147} 	imes \left( \frac{25}{81354} ight)$$\log \frac{K_{2}}{K_{1}} = 3394.78 	imes 0.000307 = 1.043$$\frac{K_{2}}{K_{1}} = 10^{1.043} \approx 11$Thus,the reaction is $11$ times faster.

How fast is the reaction at $25\, ^\circ C$ as compared to $0\, ^\circ C$,if the activation energy is $65\, kJ/mol$?

Similar Questions

For $A + B \longrightarrow C + D$; $\Delta H = -20 \ kJ \ mol^{-1}$,the activation energy of the forward reaction is $85 \ kJ \ mol^{-1}$. The activation energy for the backward reaction is.....$kJ \ mol^{-1}$

If the definition of the temperature coefficient of the reaction holds good for a reaction between $27^{\circ} C$ and $37^{\circ} C$,the activation energy for the reaction in $kJ \cdot mol^{-1}$ is

Consider the following graph of the kinetic energy distribution among molecules at temperature $T$. If the temperature is increased,how would the resulting graph differ from the one above?

Which equation is true to calculate the energy of activation,if the rate of reaction is doubled by increasing temperature from $T_1 \ K$ to $T_2 \ K$?

Which statement is incorrect according to the Arrhenius equation?

Vedclass Test Series

Exam Paper Generator

Online Exam Module