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Answer: 1. [graph option 1, see figure]

Key idea: \(\ln k=\ln A-\dfrac{E_a}{R}\cdot\dfrac1T\), a straight line with negative slope \(-E_a/R\) and positive intercept \(\ln A\).

Only graph (1) falls with increasing \(1/T\) and cuts the \(\ln k\) axis at a positive value. Graphs 2, 3 and 4 have positive slopes. Option 1.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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