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Standard XII

Chemistry

Types of Equillibrium

In the Arrhen...

Question

In the Arrhenius equation for a certain reaction, the values of $A$ and $E_{a}$ (energy of activation) are $4 \times 10^{13} s^{- 1}$ and $98.6 k J m o l^{- 1}$ , respectively. If the reaction is $1^{s t}$ order, at $X$ (kelvin) temperature and the half-life is 696 seconds, then the value of $X$ is:
(only nearest integer value)

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Solution

$k = A e^{- \frac{E_{a}}{R T}}$ $= 4 \times 10^{13} \times e^{\frac{- 98.6 \times 10^{3}}{8.314 T}}$
For $I^{s t}$ order reaction, $k = \frac{0.693}{t_{\frac{1}{2}}} = \frac{0.693}{696} = 10^{- 3} s e c^{- 1}$
$⟹ l o g 10^{- 3} = l o g (4 \times 10^{13}) - \frac{98.6 \times 10^{3}}{2.303 \times 8.314 \times T}$
$⟹ - 3 = 13.602 - \frac{98600}{19.15 T}$
$⟹ T = \frac{98600}{16.602 \times 19.15} = 310.1 K$
So, the value of $X$ is $310$ .

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In the Arrhenius equation for a certain reaction, the values of A and Ea (energy of activation) are 4×1013s−1 and 98.6 kJmol−1, respectively. If the reaction is 1st order, at X (kelvin) temperature and the half-life is 696 seconds, then the value of X is:(only nearest integer value)

k=Ae−EaRT=4×1013×e−98.6×1038.314TFor Ist order reaction, k=0.693t12=0.693696=10−3sec−1⟹log10−3=log(4×1013)−98.6×1032.303×8.314×T⟹−3=13.602−9860019.15T⟹T=9860016.602×19.15=310.1KSo, the value of X is 310.