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Chemical Kinetics question

2012 · Q65
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Chemical Kinetics question

2012 · Q65

NEETChemistryChemical KineticsMultiple correct+4 / −1
Activation energy (Eaaa) and rate constants (k1 and k2) of a chemical reaction at two different temperatures (T1 and T2) are related by
  1. A
    ln⁡k2k1=−EaR(1T1−1T2)\ln {{{k_2}} \over {{k_1}}} = - {{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)lnk1​k2​​=−REa​​(T1​1​−T2​1​)
  2. B
    ln⁡k2k1=−EaR(1T2−1T1)\ln {{{k_2}} \over {{k_1}}} = - {{{E_a}} \over R}\left( {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right)lnk1​k2​​=−REa​​(T2​1​−T1​1​)
  3. C
    ln⁡k2k1=−EaR(1T2+1T1)\ln {{{k_2}} \over {{k_1}}} = - {{{E_a}} \over R}\left( {{1 \over {{T_2}}} + {1 \over {{T_1}}}} \right)lnk1​k2​​=−REa​​(T2​1​+T1​1​)
  4. D
    ln⁡k2k1=EaR(1T1−1T2)\ln {{{k_2}} \over {{k_1}}} = {{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)lnk1​k2​​=REa​​(T1​1​−T2​1​)
View written solutionFree

Correct answer: B, D

Let k1=Ae−EaRT1{k_1} = A{e^{ - {{{E_a}} \over {R{T_1}}}}}k1​=Ae−RT1​Ea​​

ln⁡k1=ln⁡A−EaRT1\ln {k_1} = \ln A - {{{E_a}} \over {R{T_1}}}lnk1​=lnA−RT1​Ea​​ ......(1)

k2=Ae−EaRT2{k_2} = A{e^{ - {{{E_a}} \over {R{T_2}}}}}k2​=Ae−RT2​Ea​​

ln⁡k2=ln⁡A−EaRT2\ln {k_2} = \ln A - {{{E_a}} \over {R{T_2}}}lnk2​=lnA−RT2​Ea​​ ....(2)

From eq.(1) and (2), we have

ln⁡k2−ln⁡k1=ln⁡A−EaRT2−ln⁡A+EaRT1\ln {k_2} - \ln {k_1} = \ln A - {{{E_a}} \over {R{T_2}}} - \ln A + {{{E_a}} \over {R{T_1}}}lnk2​−lnk1​=lnA−RT2​Ea​​−lnA+RT1​Ea​​

⇒\Rightarrow⇒ ln⁡k2k1=EaR(1T1−1T2)\ln {{{k_2}} \over {{k_1}}} = {{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)lnk1​k2​​=REa​​(T1​1​−T2​1​)

⇒\Rightarrow⇒ ln⁡k2k1=−EaR(1T2−1T1)\ln {{{k_2}} \over {{k_1}}} = - {{{E_a}} \over R}\left( {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right)lnk1​k2​​=−REa​​(T2​1​−T1​1​)

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