The rate constant of a first order reaction becomes $5$ times when the temperature is raised from $350$ K to $400$ K. Calculate the activation energy for the reaction. (Gas reactant R $= 8.314$ J $K^{- 1} m o l^{- 1}$ ).

Open in App

Solution

Arrhenius equation is used to calculate the energy of activation of the reaction.
$l o g \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

Given that the rate constant of a first order reaction becomes 5 times when the temperature is raised from 350 K to 400 K. This means: $\frac{k_{2}}{k_{1}} = 5, T_{1} = 350 K, T_{2} = 400 K$

Substituting values in Arrhenius Equation to calculate activation energy:
$l o g \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

$l o g 5 = \frac{E_{a}}{2.303 \times 8.314 J K^{- 1} m o l^{- 1}} [\frac{1}{350 K} - \frac{1}{400 K}]$

$l o g 5 = \frac{E_{a}}{2.303 \times 8.314 J K^{- 1} m o l^{- 1}} [\frac{400 K - 350 K}{350 K \times 400 K}]$

$0.699 = \frac{E_{a}}{2.303 \times 8.314 J m o l^{- 1}} [\frac{50 K}{140000 K}]$

$E_{a} = \frac{0.699 \times 2.303 \times 8.314 J m o l^{- 1} \times 140000}{50}$

$E_{a} = 37474.8 J m o l^{- 1}$

$E_{a} = 37474.8 J m o l^{- 1} \times \frac{1.00 k J}{1000 J} = 37.5 k J m o l^{- 1}$

Suggest Corrections

Similar questions

350 K

400 K

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

30^{0} C

50^{0} C

J m o l^{- 1} K^{- 1}

293 K

313 K

(E_{2})

[R = 8.314 J / K m o l^{- 1}, log 4 = 0.6021]

m o l^{- 1}

K^{- 1}

20^{0} C

35^{0} C

m o l^{- 1} K^{- 1}

Join BYJU'S Learning Program