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  • In Arrhenius equation for a certain reaction, the value of A (Arrhenius factor) and <img alt="\mathrm{E}_{\mathrm{a}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BE%7D_%7B%5Cma

In Arrhenius equation for a certain reaction, the value of A (Arrhenius factor) and \mathrm{E}_{\mathrm{a}} (activation energy) are 4 \times 10^{13} \mathrm{~s}^{-1} and 98.6 \mathrm{~kJ} \mathrm{~mol}^{-1} respectively. At what temperature, the reaction will have specific rate constant of 1.1 \times 10^{-3} \mathrm{~s}^{-1} ?
Take
\log \left(\frac{1.1}{4}\right)=-0.56

Option: 1

79 K


Option: 2

34 K


Option: 3

54 K


Option: 4

311 K


Answers (1)

According to Arrhenius the relationship between temperature and rate constant is given by,
\mathrm{k=A e^{-E_a / R T}}
where,
K : Rate constant
A : Arrhenius factor or frequency factor or pre-exponential factor
\mathrm{E_a} : Activation energy (in $\mathrm{J} / \mathrm{mol}$ )
R : Gas constant
T : Temperature of reaction (in Kelvin(K))
\mathrm{k=A e^{-E_n / R T}}
Taking In on both side
\mathrm{\ln k=\ln A-\frac{E_a}{R T}}
\mathrm{2.303 \log k=2.303 \log A-\frac{E_a}{R T}}
\mathrm{2.303 \log \left(1.1 \times 10^{-3}\right)=2.303 \log \left(4 \times 10^{13}\right)-\frac{98.6 \times 10^3}{8.314 \times T}}
\mathrm{2.303\left(\log \frac{1.1 \times 10^{-3}}{4 \times 10^{13}}\right)=\frac{98.6 \times 10^3}{8.314 \times T}}
\mathrm{2.303\left(\log \left(\frac{1.1}{4}\right)-16\right)=-\frac{98.6 \times 10^3}{8.314 \times T}}
\mathrm{-2.303 \times 16.56=-\frac{98.6 \times 10^3}{8.314 \times T}}
\mathrm{ T=\frac{98.6 \times 10^3}{8.314 \times 2.303 \times 16.56} }

\mathrm{ =310.96 \mathrm{~K}}

Posted by

Ramraj Saini

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