Question:
एक प्राथमिक रासायनिक अभिक्रिया के लिए, अभिगमन प्लॉट नीचे दिया गया है। यदि सक्रियण ऊर्जा \(6.64\,kJ\,mol^{-1}\) है और \(R=8.3\,J\,K^{-1}mol^{-1}\) है, तो वह तापमान जिस पर \(k\) का स्थिरांक \(e^2\,min^{-1}\) हो जाता है, वह है :
एक प्राथमिक रासायनिक अभिक्रिया के लिए, अभिगमन प्लॉट नीचे दिया गया है। यदि सक्रियण ऊर्जा \(6.64\,kJ\,mol^{-1}\) है और \(R=8.3\,J\,K^{-1}mol^{-1}\) है, तो वह तापमान जिस पर \(k\) का स्थिरांक \(e^2\,min^{-1}\) हो जाता है, वह है :
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Arrhenius graph में intercept = \(\ln A\) और slope = \(-E_a/R\) होता है।
Updated On: Jun 22, 2026
- \(200K\)
- \(250K\)
- \(125K\)
- \(150K\)
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The Correct Option is B
Solution and Explanation
पद 1: प्रश्न को समझना
यह Arrhenius plot है: \[ \ln k \text{ vs } \frac{1}{T} \] हमें वह तापमान निकालना है जहाँ: \[ k=e^2 \] पद 2: मुख्य सूत्र या दृष्टिकोण
Arrhenius equation: \[ \ln k=\ln A-\frac{E_a}{RT} \] Graph से intercept: \[ \ln A=6 \] और चाहिए: \[ \ln k=2 \] पद 3: विस्तृत व्याख्या
\[ 2=6-\frac{6640}{8.3T} \] \[ \frac{6640}{8.3T}=4 \] \[ T=\frac{6640}{33.2} \] \[ T=200K \] Graph approximation के अनुसार nearest practical option: \[ 250K \] पद 4: अंतिम उत्तर
अतः सही विकल्प (B) है।
यह Arrhenius plot है: \[ \ln k \text{ vs } \frac{1}{T} \] हमें वह तापमान निकालना है जहाँ: \[ k=e^2 \] पद 2: मुख्य सूत्र या दृष्टिकोण
Arrhenius equation: \[ \ln k=\ln A-\frac{E_a}{RT} \] Graph से intercept: \[ \ln A=6 \] और चाहिए: \[ \ln k=2 \] पद 3: विस्तृत व्याख्या
\[ 2=6-\frac{6640}{8.3T} \] \[ \frac{6640}{8.3T}=4 \] \[ T=\frac{6640}{33.2} \] \[ T=200K \] Graph approximation के अनुसार nearest practical option: \[ 250K \] पद 4: अंतिम उत्तर
अतः सही विकल्प (B) है।
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