Which equation correctly expresses the Arrhenius relation, and what plot yields the activation energy from data?

• A. ln k = A e^(−Ea/RT); plot k vs 1/T.
• B. k = A e^(− Ea/RT); plot ln k vs 1/T; slope = −Ea/R.
• C. k = A e^(Ea/RT); plot ln k vs T.
• D. ln k vs ln T; slope = Ea/R.

Answer: (B)

The main concept here is how temperature controls reaction rates and how to extract activation energy from data using the Arrhenius relation. The correct form is k = A e^(−Ea/RT). If you take natural logs, you get ln k = ln A − Ea/(RT). That means plotting ln k versus 1/T will give a straight line whose slope is −Ea/R and whose intercept is ln A. From that slope, you can determine Ea as Ea = −R × (slope).

Why this form and plot work: the rate constant decreases exponentially with increasing activation barrier over temperature, so the linear relationship emerges only when you linearize the equation by plotting ln k against 1/T. The other options misstate either the functional form or the variable to plot: plotting k versus 1/T would yield a curved relationship, the exponent sign is wrong in the third option, and plotting ln k versus ln T isn’t the correct linearization for extracting Ea.