In Arrhenius equation for a certain reaction- the values of A and E a activation energy are 4- 10 13 sec -1 and 98-6 kJ mol -1 respectively- If the reaction is of first order- at what temperature will its half life period be 10 minutes-

According to Arrhenius equation, k=A e ^ E _ a / RT or log _ e k =log _ e A E _ a RT or 2.303log _ 10 k =2.303log _ 10 A E _ a RT ...

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Standard XII

Chemistry

Arrhenius Equation

In Arrhenius ...

Question

In Arrhenius equation for a certain reaction, the values of $A$ and $E_{a}$ (activation energy) are $4 \times 10^{13} {s e c}^{- 1}$ and $98.6$ $k J$ ${m o l}^{- 1}$ respectively. If the reaction is of first order, at what temperature will its half life period be $10$ minutes?

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Solution

According to Arrhenius equation,
$k = A e^{- E_{a} / R T}$
or ${log}_{e} k = {log}_{e} A - \frac{E_{a}}{R T}$
or $2.303 {log}_{10} k = 2.303 {log}_{10} A - \frac{E_{a}}{R T}$
For a first order reaction $t_{1 / 2} = \frac{0.693}{k}$
So, $k = \frac{0.693}{600} {s e c}^{- 1} (t_{1 / 2} = 10 m i n = 600 s e c)$
$= 1.1 \times 10^{- 3} {s e c}^{- 1}$
Hence, $log (1.1 \times 10^{- 3}) = log (4 \times 10^{13}) - \frac{98.6 \times 10^{3}}{2.303 \times 8.314 \times T}$
$T = 310.95$ $K$

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In Arrhenius equation for a certain reaction, the values of A and Ea (activation energy) are 4×1013sec−1 and 98.6 kJ mol−1 respectively. If the reaction is of first order, at what temperature will its half life period be 10 minutes?

According to Arrhenius equation,k=Ae−Ea/RTor logek=logeA−EaRTor 2.303log10k=2.303log10A−EaRTFor a first order reaction t1/2=0.693kSo, k=0.693600sec−1(t1/2=10min=600sec) =1.1×10−3sec−1Hence, log(1.1×10−3)=log(4×1013)−98.6×1032.303×8.314×TT=310.95 K

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