Chemistry

Chemistry

Problem

For a reaction A2+B2⟶2ABA_2 + B_2 \longrightarrow 2AB , evaluate the energy of activation from the following data :
TT (in K)1T\frac{1}{T} ( K−1K^{-1} )log⁡10K\log_{10} K
5002×10−32 \times 10^{-3}3.0
2005×10−35 \times 10^{-3}2.0
Intuitive explanation

The activation energy is hidden in how the rate constant changes with temperature—this is the Arrhenius equation at work. By plotting log⁡10K\log_{10} K against 1T\frac{1}{T}, you get a straight line whose slope directly reveals EaE_a. The key is recognizing that the slope equals Ea2.303R\frac{E_a}{2.303R}, so you can extract activation energy from just two data points without needing to know the pre-exponential factor.

Intuitive explanation

The activation energy is hidden in how the rate constant changes with temperature—this is the Arrhenius equation at work. By plotting log⁡10K\log_{10} K against 1T\frac{1}{T}, you get a straight line whose slope directly reveals EaE_a. The key is recognizing that the slope equals Ea2.303R\frac{E_a}{2.303R}, so you can extract activation energy from just two data points without needing to know the pre-exponential factor.

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