When ln k is plotted against 1 - T- the slope was found to be -10-7 - 10 3 K- activation energy for the reaction would be -

The correct option is C 88.9 kJ mol ^ 1 From the Arrhenius equation, we get lnk=lnA E_aRT Given, E_aR= 10.7×10^3 ⇒ E_a=8.314×10.7×10^3 Jmol ...

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Standard XII

Chemistry

Activation Energy

When ln k is ...

Question

When ln k is plotted against $1 / T$ , the slope was found to be -10.7 $\times$ $10^{3}$ K, activation energy for the reaction would be :

-78.9 kJ

{m o l}^{- 1}

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2.26 kJ

{m o l}^{- 1}

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88.9 kJ

{m o l}^{- 1}

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10.7 kJ

{m o l}^{- 1}

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Solution

The correct option is C 88.9 kJ ${m o l}^{- 1}$
From the Arrhenius equation, we get
$l n k = l n A - \frac{E_{a}}{R T}$
Given,
$- \frac{E_{a}}{R} = - 10.7 \times 10^{3}$

$\Rightarrow E_{a} = 8.314 \times 10.7 \times 10^{3} J m o l^{- 1}$
$\Rightarrow E_{a} = 88.9 k J m o l^{- 1}$

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When ln k is plotted against 1/T, the slope was found to be -10.7 × 103 K, activation energy for the reaction would be :

The correct option is C 88.9 kJ mol−1From the Arrhenius equation, we getlnk=lnA−EaRTGiven,−EaR=−10.7×103⇒Ea=8.314×10.7×103 Jmol−1⇒Ea=88.9 kJmol−1

When ln k is plotted against $1 / T$ , the slope was found to be -10.7 $\times$ $10^{3}$ K, activation energy for the reaction would be :

The correct option is C 88.9 kJ ${m o l}^{- 1}$
From the Arrhenius equation, we get
$l n k = l n A - \frac{E_{a}}{R T}$
Given,
$- \frac{E_{a}}{R} = - 10.7 \times 10^{3}$

$\Rightarrow E_{a} = 8.314 \times 10.7 \times 10^{3} J m o l^{- 1}$
$\Rightarrow E_{a} = 88.9 k J m o l^{- 1}$