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Write Brief Answer · Q27

Q.The activation energy of a reaction is 22.5 k Cal mol−122.5\ \text{k Cal mol}^{-1} and the value of rate constant at 40∘C40^\circ C is 1.8×10−5 s−11.8\times10^{-5}\ s^{-1}. Calculate the frequency factor, A.

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Step 1. T=40∘C+273=313T=40^\circ\text{C}+273=313 K; Ea=22.5 kCal mol−1=22,500 cal mol−1E_a=22.5\ \text{kCal mol}^{-1}=22{,}500\ \text{cal mol}^{-1}; using R=1.987 cal K−1mol−1R=1.987\ \text{cal K}^{-1}\text{mol}^{-1} (to match Ea's calorie units).

Step 2. Ea2.303RT=22,5002.303×1.987×313=22,5001432.6≈15.71\dfrac{E_a}{2.303RT}=\dfrac{22{,}500}{2.303\times1.987\times313}=\dfrac{22{,}500}{1432.6}\approx15.71.

Step 3. log⁡k=log⁡(1.8×10−5)=log⁡1.8−5=0.2553−5=−4.7447\log k=\log(1.8\times10^{-5})=\log1.8-5=0.2553-5=-4.7447.

Step 4. log⁡A=log⁡k+Ea2.303RT=−4.7447+15.71=10.96⇒A=1010.96≈9.1×1010 s−1\log A=\log k+\dfrac{E_a}{2.303RT}=-4.7447+15.71=10.96\Rightarrow A=10^{10.96}\approx9.1\times10^{10}\ \text{s}^{-1}. …

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