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Temperature at Which Rate Constant Increases using Arrhenius Equation

For reaction A→P\mathrm{A \rightarrow P}, rate constant k=1.5×103 s−1\mathrm{k} = 1.5 \times 10^3\ \text{s}^{-1} at 27∘C27^\circ\text{C} If activation energy for the above reaction is 60 kJ mol−160\text{ kJ mol}^{-1}, then the temperature (in ∘C^\circ\text{C}) at which rate constant, k=4.5×103 s−1\mathrm{k} = 4.5 \times 10^3\ \text{s}^{-1} is ________. (Nearest integer) Given : log⁡2=0.30\log 2 = 0.30, log⁡3=0.48\log 3 = 0.48, R=8.3 J K−1mol−1\mathrm{R} = 8.3\text{ J K}^{-1}\text{mol}^{-1}, ln⁡10=2.3\ln 10 = 2.3

Official Numerical Answer41

Topics & Concepts

Chemical KineticsArrhenius EquationRate of Reaction

Step-by-Step Solution

To find the temperature at which the rate constant increases to k2=4.5×103 s−1k_2 = 4.5 \times 10^3\ \text{s}^{-1}, we use the integrated form of the Arrhenius equation:

ln⁡(k2k1)=EaR(1T1−1T2)\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)

Given parameters:

  • Initial rate constant, k1=1.5×103 s−1k_1 = 1.5 \times 10^3\ \text{s}^{-1}
  • Initial temperature, T1=27∘C=27+273=300 KT_1 = 27^\circ\text{C} = 27 + 273 = 300\text{ K}
  • Final rate constant, k2=4.5×103 s−1k_2 = 4.5 \times 10^3\ \text{s}^{-1}
  • Activation energy, Ea=60 kJ mol−1=60×103 J mol−1E_a = 60\text{ kJ mol}^{-1} = 60 \times 10^3\text{ J mol}^{-1}
  • Universal gas constant, R=8.3 J K−1mol−1R = 8.3\text{ J K}^{-1}\text{mol}^{-1}
  • ln⁡10=2.3\ln 10 = 2.3
  • log⁡3=0.48\log 3 = 0.48

We can write ln⁡(k2k1)\ln\left(\frac{k_2}{k_1}\right) in terms of base-10 logarithm as: ln⁡(k2k1)=(ln⁡10)×log⁡(k2k1)\ln\left(\frac{k_2}{k_1}\right) = (\ln 10) \times \log\left(\frac{k_2}{k_1}\right)

Substitute the ratio of rate constants: k2k1=4.5×1031.5×103=3\frac{k_2}{k_1} = \frac{4.5 \times 10^3}{1.5 \times 10^3} = 3

Thus, the Arrhenius equation becomes: (ln⁡10)×log⁡(3)=EaR(1T1−1T2)(\ln 10) \times \log(3) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)

Substitute the given values into the equation: 2.3×0.48=60×1038.3(1300−1T2)2.3 \times 0.48 = \frac{60 \times 10^3}{8.3} \left(\frac{1}{300} - \frac{1}{T_2}\right)

1.104=600008.3(1300−1T2)1.104 = \frac{60000}{8.3} \left(\frac{1}{300} - \frac{1}{T_2}\right)

Rearranging to solve for (1300−1T2)\left(\frac{1}{300} - \frac{1}{T_2}\right): 1300−1T2=1.104×8.360000\frac{1}{300} - \frac{1}{T_2} = \frac{1.104 \times 8.3}{60000}

1300−1T2=9.163260000\frac{1}{300} - \frac{1}{T_2} = \frac{9.1632}{60000}

1T2=1300−9.163260000=200−9.163260000=190.836860000\frac{1}{T_2} = \frac{1}{300} - \frac{9.1632}{60000} = \frac{200 - 9.1632}{60000} = \frac{190.8368}{60000}

Solving for T2T_2: T2=60000190.8368≈314.4 KT_2 = \frac{60000}{190.8368} \approx 314.4\text{ K}

To find the temperature in ∘C^\circ\text{C}: t2=T2−273=314.4−273=41.4∘Ct_2 = T_2 - 273 = 314.4 - 273 = 41.4^\circ\text{C}

Rounding off to the nearest integer gives 4141.

Temperature at Which Rate Constant Increases using Arrhenius Equation | Chemistry PYQ Solution - JEE Challenger