The activation energy of the reaction:
$A + B \to$ Products is 105.73 KJ/mol. At $40^{\circ} C$ , the products are formed at the rate of 0.133 mol $L^{- 1} m i n^{- 1}$ the rate of formation of products at $80^{\circ} C$ will be

13.3 m o l L^{- 1} m i n^{- 1}

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5.6 m o l L^{- 1} m i n^{- 1}

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17.8 m o l L^{- 1} m i n^{- 1}

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Solution

The correct option is A $13.3 m o l L^{- 1} m i n^{- 1}$
Let the rate law be defined as
$A t T_{1} : r_{1} = k_{1} [A]^{x} [B]^{y}$
$A t T_{2} : r_{2} = k_{2} [A]^{x} [B]^{y}$
$\Rightarrow r_{2} = r_{1} [\frac{k_{2}}{k_{1}}]$
Using Arrhenius equation, find k at $40^{\circ} C$
$l o g \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$
$\Rightarrow l o g \frac{k_{2}}{k_{1}} = \frac{105.73 \times 10^{3}}{2.303 \times 8.31} (\frac{40}{313 \times 353})$
$\Rightarrow l o g \frac{k_{2}}{k_{1}} = 2.0$
$\Rightarrow \frac{k_{2}}{k_{1}} = 100$
$\Rightarrow r_{2} = 0.133 \times 100 = 13.3 m o l L^{- 1} m i n^{- 1}$

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