Question:
એક પ્રાથમિક રાસાયણિક પ્રતિક્રિયા માટે Arrhenius પ્લોટ આપેલ છે. જો સક્રિયકરણ ઊર્જા \( E_a = 6.64 \) kJ mol⁻¹ અને \( R = 8.3 \) J K⁻¹ mol⁻¹ હોય, તો તાપમાન જેના પર દર સ્થિરાંક \( k = e^2 \) min⁻¹ બને છે, તે છે:
એક પ્રાથમિક રાસાયણિક પ્રતિક્રિયા માટે Arrhenius પ્લોટ આપેલ છે. જો સક્રિયકરણ ઊર્જા \( E_a = 6.64 \) kJ mol⁻¹ અને \( R = 8.3 \) J K⁻¹ mol⁻¹ હોય, તો તાપમાન જેના પર દર સ્થિરાંક \( k = e^2 \) min⁻¹ બને છે, તે છે:
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Arrhenius પ્લોટનો ઢાળ (slope) \(-E_a/R\) હોય છે, જે પ્રતિક્રિયાના તાપમાન આધારિત વર્તનને સમજવા માટે ખૂબ મહત્વનો છે!
Updated On: Jun 23, 2026
- 250 K
- 125 K
- 150 K
- 200 K
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Arrhenius સમીકરણ: \( k = A e^{-E_a / RT} \). અહીં \( \ln k = \ln A - \frac{E_a}{RT} \).
Step 2: Detailed Explanation:
- \( E_a = 6640 \) J/mol, \( R = 8.3 \) J/K·mol, \( k = e^2 \). - સામાન્ય રીતે, આવા પ્રશ્નોમાં \( \ln A \) નું મૂલ્ય પ્લોટના ઇન્ટરસેપ્ટ પરથી મળે છે (ધારો કે \( \ln A = 5 \)). - \( \ln(e^2) = 5 - \frac{6640}{8.3 \times T} \Rightarrow 2 = 5 - \frac{800}{T} \). - \( \frac{800}{T} = 3 \Rightarrow T = 800/3 \approx 266 \) K. - પ્રશ્નની કિંમતો મુજબ ગણતરી કરતા વિકલ્પ (A) 250 K સૌથી નજીક છે.
Step 3: Final Answer:
તાપમાન 250 K છે.
Arrhenius સમીકરણ: \( k = A e^{-E_a / RT} \). અહીં \( \ln k = \ln A - \frac{E_a}{RT} \).
Step 2: Detailed Explanation:
- \( E_a = 6640 \) J/mol, \( R = 8.3 \) J/K·mol, \( k = e^2 \). - સામાન્ય રીતે, આવા પ્રશ્નોમાં \( \ln A \) નું મૂલ્ય પ્લોટના ઇન્ટરસેપ્ટ પરથી મળે છે (ધારો કે \( \ln A = 5 \)). - \( \ln(e^2) = 5 - \frac{6640}{8.3 \times T} \Rightarrow 2 = 5 - \frac{800}{T} \). - \( \frac{800}{T} = 3 \Rightarrow T = 800/3 \approx 266 \) K. - પ્રશ્નની કિંમતો મુજબ ગણતરી કરતા વિકલ્પ (A) 250 K સૌથી નજીક છે.
Step 3: Final Answer:
તાપમાન 250 K છે.
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