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- At what temperature- the reaction will have specific rate constant 1-1- 10 -3 sec -1-

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

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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. At what temperature, the reaction will have specific rate constant $1.1 \times 10^{- 3} {s e c}^{- 1}$ ?

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Solution

According to Arrhenius equation,
$k = A e^{- E_{a} / R T}$
or ${log}_{e} k = {log}_{e} A - \frac{E_{a}}{R T} {log}_{e} e$
or $2.303 {log}_{10} k = 2.303 {log}_{10} A - \frac{E_{a}}{R T}$
or $2.303 log (1.1 \times 10^{- 3}) = 2.303 log (4 \times 10^{13}) - \frac{98.6 \times 10^{3}}{8.314 \times T}$
$T = \frac{98.6 \times 10^{3}}{8.314 \times 2.303 \times 16.56} K = 310.96 K$

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In Arrhenius equation for a certain reaction, the values of A and Ea (activation energy) are 4×1013 sec−1 and 98.6 kJ mol−1 respectively. At what temperature, the reaction will have specific rate constant 1.1×10−3 sec−1?

According to Arrhenius equation,k=Ae−Ea/RTor logek=logeA−EaRTlogeeor 2.303log10k=2.303log10A−EaRTor 2.303log(1.1×10−3)=2.303log(4×1013)−98.6×1038.314×TT=98.6×1038.314×2.303×16.56K=310.96K

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. At what temperature, the reaction will have specific rate constant $1.1 \times 10^{- 3} {s e c}^{- 1}$ ?