JEE Main 2023 · 8 April · morning · Numerical answer · +4 / −0
The number of given statement/s which is/are correct is __________.
(A) The stronger the temperature dependence of the rate constant, the higher is the activation energy.
(B) If a reaction has zero activation energy, its rate is independent of temperature.
(C) The stronger the temperature dependence of the rate constant, the smaller is the activation energy.
(D) If there is no correlation between the temperature and the rate constant then it means that the reaction has negative activation energy.
Answer and solution
Answer 2
Solution
Chemistry ma’am · 3 steps
Read every statement off the ln k versus 1/T line, whose slope is -Eₐ/R.
ln k = ln A - Eₐ/R·1/T
- Slope is Ea. A plot of ln k against 1/T is a straight line with slope -Eₐ/R. A steeper line means k changes more with temperature. So a stronger temperature dependence means a larger Eₐ.
- Zero activation energy. If Eₐ = 0, the exponential term becomes e⁰ = 1 and k = A. The line is flat, so the rate constant does not change with temperature.
Statement What the equation says Verdict (A) Steeper slope means larger Eₐ Correct (B) Eₐ = 0 gives k = A, a flat line Correct (C) The opposite of (A) Incorrect (D) No change of k with T means zero slope, so Eₐ = 0, not a negative value Incorrect - So. Only (A) and (B) are correct, so the count is 2.
The common slip: 3. Counting (D) as correct gives 3. When k does not change with temperature, the slope is zero, which means Eₐ = 0, not a negative value.
The idea to keep. Plot ln k against 1/T: the slope is -Eₐ/R. A steep line means a large Eₐ and a flat line means Eₐ = 0. A negative Eₐ would make k fall as T rises.
Answer from the NTA final answer key. Solution by PrepWiser.