
The answer is D.
• The activation energy of a reaction does not change with temperature, so the new intercept must still lie on the dashed \(E_\text{A}\) line. This rules out A and C, which are not on the \(E_\text{A}\) line.
• At the lower temperature, \((T - 50)^\circ\text{C}\), the peak of the Boltzmann distribution moves to lower energy and the curve at high energies falls. Fewer particles have energy equal to or greater than \(E_\text{A}\), so the new curve crosses the \(E_\text{A}\) line lower than the present curve does.
• D lies on the \(E_\text{A}\) line below the present intercept, so D is correct.
• B is wrong: it lies on the \(E_\text{A}\) line above the present intercept, which would mean more particles at \(E_\text{A}\), as happens when the temperature is raised.
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