The slope of Arrhenius plot ( ln k vs 1 T ) of first order reaction is - 5 × 10 3 K . The value of E a of the reaction is [Given : R = 8 . 314 J K - 1 mol

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Q.
The slope of Arrhenius plot $(\ln k vs \frac{1}{T})$ of first order reaction is $- 5 \times 10^{3}$ $K$ . The value of $E_{a}$ of the reaction is [Given : $R = 8.314$ $J$ $K^{- 1}$ ${mol}^{- 1}$ ] [2021]

1 $- 83$ $kJ$ ${mol}^{- 1}$

2 $41.5$ $kJ$ ${mol}^{- 1}$

3 $83.0$ $kJ$ ${mol}^{- 1}$

4 $166$ $kJ$ ${mol}^{- 1}$

Ans.
(2)

$\ln k = - \frac{E_{a}}{R T} + \ln A$

$- \frac{E_{a}}{R} = - 5 \times 10^{3} = - 5000$

$\Rightarrow E_{a} = 5000 \times 8.314 = 41570$ $J$ ${mol}^{- 1} = 41.57$ $kJ$ ${mol}^{- 1}$

Solution

Q.
The slope of Arrhenius plot $(\ln k vs \frac{1}{T})$ of first order reaction is $- 5 \times 10^{3}$ $K$ . The value of $E_{a}$ of the reaction is [Given : $R = 8.314$ $J$ $K^{- 1}$ ${mol}^{- 1}$ ] [2021]

1 $- 83$ $kJ$ ${mol}^{- 1}$

2 $41.5$ $kJ$ ${mol}^{- 1}$

3 $83.0$ $kJ$ ${mol}^{- 1}$

4 $166$ $kJ$ ${mol}^{- 1}$

Ans.
(2)

$\ln k = - \frac{E_{a}}{R T} + \ln A$

$- \frac{E_{a}}{R} = - 5 \times 10^{3} = - 5000$

$\Rightarrow E_{a} = 5000 \times 8.314 = 41570$ $J$ ${mol}^{- 1} = 41.57$ $kJ$ ${mol}^{- 1}$

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Solution

Q. The slope of Arrhenius plot (lnk vs 1T) of first order reaction is -5×103 K. The value of Ea of the reaction is [Given :R=8.314 J K-1 mol-1] [2021]

1 -83 kJ mol-1

2 41.5 kJ mol-1

3 83.0 kJ mol-1

4 166 kJ mol-1

Ans. (2) lnk=-EaRT+lnA -EaR=-5×103=-5000 ⇒ Ea=5000×8.314=41570 J mol-1=41.57 kJ mol-1

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Q.
The slope of Arrhenius plot $(\ln k vs \frac{1}{T})$ of first order reaction is $- 5 \times 10^{3}$ $K$ . The value of $E_{a}$ of the reaction is [Given : $R = 8.314$ $J$ $K^{- 1}$ ${mol}^{- 1}$ ] [2021]

1 $- 83$ $kJ$ ${mol}^{- 1}$

2 $41.5$ $kJ$ ${mol}^{- 1}$

3 $83.0$ $kJ$ ${mol}^{- 1}$

4 $166$ $kJ$ ${mol}^{- 1}$

Ans.
(2)

$\ln k = - \frac{E_{a}}{R T} + \ln A$

$- \frac{E_{a}}{R} = - 5 \times 10^{3} = - 5000$

$\Rightarrow E_{a} = 5000 \times 8.314 = 41570$ $J$ ${mol}^{- 1} = 41.57$ $kJ$ ${mol}^{- 1}$