Verified Solution • 2025–26 Session moderate

QAIPM-2012-Ch-1631: NEET 2012 Chemistry - Class 12 AIPM-2012-Ch-1631 Solution: Activation energy (\(E_a\)) and rate constants (\(k_1\) and…

Activation energy (EaE_a) and rate constants (k1k_1 and k2k_2) of a chemical reaction at two different temperatures (T1T_1 and T2T_2) are related by ln⁡k2k1=−EaR(1T1−1T2)\ln\frac{k_2}{k_1} = -\frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2}) ln⁡k2k1=−EaR(1T2−1T1)\ln\frac{k_2}{k_1} = -\frac{E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1}) ln⁡k2k1=−EaR(1T2+1T1)\ln\frac{k_2}{k_1} = -\frac{E_a}{R}(\frac{1}{T_2}+\frac{1}{T_1}) ln⁡k2k1=EaR(1T1−1T2)\ln\frac{k_2}{k_1} = \frac{E_a}{R}(\frac{1}{T_1}-\frac{1}{T_2})

Multiple Choice Options

  • 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) ln k 1 _ k 2 _ _ = _ R E a _ _ ( T 1 _ 1 _ _ T 2 _ 1 _ )
  • 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) ln k 1 _ k 2 _ _ = _ R E a _ _ ( T 2 _ 1 _ _ T 1 _ 1 _ )
  • 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) ln k 1 _ k 2 _ _ = _ R E a _ _ ( T 2 _ 1 _ + T 1 _ 1 _ )
  • 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) ln k 1 _ k 2 _ _ = R E a _ _ ( T 1 _ 1 _ _ T 2 _ 1 _ )

Step-by-Step Verified Solution

Let k 1 = A e _ E a R T 1 ln _ k 1 = ln _ A _ E a R T 1 ......(1) k 2 = A e _ E a R T 2 ln _ k 2 = ln _ A _ E a R T 2 ....(2) From eq.(1) and (2), we have ln _ k 2 _ ln _ k 1 = ln _ A _ E a R T 2 _ ln _ A + E a R T 1 _ ln _ k 2 k 1 = E a R ( 1 T 1 _ 1 T 2 ) _ ln _ k 2 k 1 = _ E a R ( 1 T 2 _ 1 T 1 )

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