The rate of a particular reaction doubles when the temperature changes from $300 K$ to $310 K$ . Calculate the energy of activation of the reaction. [Given $R = 8.314 J K^{- 1} {mol}^{- 1}$ ]

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Solution

Arrhenius equation
The Arrhenius equation can be used to determine the influence of temperature change on the rate constant and the rate of the reaction.
The Arrhenius equation is as follows:
$\log (\frac{k_{2}}{k_{1}}) = \frac{E_{a}}{2.303 R} (\frac{T_{2} - T_{1}}{T_{2} T_{1}})$
$k_{2}$ and $k_{1}$ are the rate constants.
$T_{2}$ and $T_{1}$ are the temperatures.
$E_{a}$ is the activation energy.
Determine activation energy
Initial temperature, $T_{1} = 300 K$
Final temperature, $T_{2} = 310 K$
Initial rate constant is $k_{1}$ .
Final rate constant, $k_{2} = 2 k_{1}$
The activation energy is as follows:
$\begin{array}{rcl} \log (\frac{k_{2}}{k_{1}}) & = & \frac{E_{a}}{2.303 R} (\frac{T_{2} - T_{1}}{T_{2} T_{1}}) \\ E_{a} & = & \log (\frac{k_{2}}{k_{1}}) \times 2.303 R \times (\frac{T_{2} T_{1}}{T_{2} - T_{1}}) \\ = & \log (\frac{2 k_{1}}{k_{1}}) \times 2.303 \times 8.314 J K^{- 1} {mol}^{- 1} \times (\frac{310 K \times 300 K}{310 K - 300 K}) \\ = & 0.3010 \times 19.15 J K^{- 1} {mol}^{- 1} \times 9300 K \\ = & 53606.60 J {mol}^{- 1} \\ = & 53.60 kJ {mol}^{- 1} \end{array}$
The activation energy of the reaction is $\begin{array}{rcl} 53.60 kJ {mol}^{- 1} \end{array}$ .

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Similar questions

Q. a) The rate of a particular reaction doubles when the temperature changes from 300 K to 310 K. Calculate the energy of activation of the reaction.
[Given :

R = 8.314 J K^{- 1} {m o l}^{- 1}

].
b) Show that the half-life period of a first order reaction is independent of initial concentration of reacting species.

Q. The rate of a reaction doubles when its temperature changes from

300 K

310 K

. Activation energy of such a reaction will be:

[R = 8.314 J K^{- 1} m o l^{- 1}

and

log 2 = 0.301]

Q. The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be:

(R = 8.314 J K^{- 1} m o l^{- 1}

and

l o g_{10} 2 = 0.301)

The rate of a reaction doubles when its temperature changes form 300 K to 310 K. Activation energy of such a reaction will be $(R = 8.314 J K^{- 1} m o l^{- 1} a n d l o g 2 = 0.301)$

The rate of reaction doubles when its temperature changes from 300K to 310 K .activation energy of such a reaction will be (R=8.314 kj/ mol and WG2= 0.031 )

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The rate of a particular reaction doubles when the temperature changes from 300K to 310K. Calculate the energy of activation of the reaction. [Given R=8.314JK−1mol−1 ]

The rate of a particular reaction doubles when the temperature changes from $300 K$ to $310 K$ . Calculate the energy of activation of the reaction. [Given $R = 8.314 J K^{- 1} {mol}^{- 1}$ ]