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ChemMedMain 2026 • 8th Apr - Shift 2

Activation Energy Temperature Coefficient and Half Life Graph

Given below are two statements :

R=8.314 J K−1 mol−1R = 8.314\text{ J K}^{-1}\text{ mol}^{-1} and 1 cal=4.2 J1\text{ cal} = 4.2\text{ J}

Statement I : When Ea=12.6 kcal/molE_a = 12.6\text{ kcal/mol}, the room temperature rate constant is doubled by a 10∘C10^{\circ}\text{C} increase in temperature (298 K298\text{ K} to 308 K308\text{ K})

Statement II : For a first order reactions A→B\text{A} \rightarrow \text{B},

Here [A]o[\text{A}]_o is the initial concentration of A\text{A} and t1/2t_{1/2} is half life of reaction. In the light of the above statements, choose the correct answer from the options given below :

Question Diagram 1

Options

A

Both Statement I and Statement II are true

B

Both Statement I and Statement II are false

C

Statement I is true but Statement II is false

Correct
D

Statement I is false but Statement II is true

Topics & Concepts

Chemical KineticsArrhenius EquationOrder of Reaction

Step-by-Step Solution

To determine the correctness of Statement I and Statement II, let us evaluate each statement individually.

Analysis of Statement I:

According to the Arrhenius equation, the temperature dependence of the rate constant kk is given by: ln⁡(k2k1)=EaR(T2−T1T1T2)\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{T_2 - T_1}{T_1 T_2}\right)

Given parameters:

  • T1=298 KT_1 = 298\text{ K}
  • T2=308 KT_2 = 308\text{ K}
  • k2k1=2\frac{k_2}{k_1} = 2 (since the rate constant is doubled)
  • R=8.314 J K−1 mol−1R = 8.314\text{ J K}^{-1}\text{ mol}^{-1} and 1 cal=4.2 J1\text{ cal} = 4.2\text{ J} R=8.3144.2 cal K−1 mol−1≈1.9795×10−3 kcal K−1 mol−1R = \frac{8.314}{4.2}\text{ cal K}^{-1}\text{ mol}^{-1} \approx 1.9795 \times 10^{-3}\text{ kcal K}^{-1}\text{ mol}^{-1}

Substituting these values into the equation: ln⁡(2)=Ea1.9795×10−3 kcal K−1 mol−1×(308−298298×308)\ln(2) = \frac{E_a}{1.9795 \times 10^{-3}\text{ kcal K}^{-1}\text{ mol}^{-1}} \times \left(\frac{308 - 298}{298 \times 308}\right)

0.69315=Ea1.9795×10−3×10917840.69315 = \frac{E_a}{1.9795 \times 10^{-3}} \times \frac{10}{91784}

Ea=0.69315×1.9795×10−3×9178410E_a = \frac{0.69315 \times 1.9795 \times 10^{-3} \times 91784}{10}

Ea≈12.59 kcal/mol≈12.6 kcal/molE_a \approx 12.59\text{ kcal/mol} \approx 12.6\text{ kcal/mol}

Since the calculated activation energy matches 12.6 kcal/mol12.6\text{ kcal/mol}, Statement I is true.

Analysis of Statement II:

For a first-order reaction A→B\text{A} \rightarrow \text{B}, the expression for half-life (t1/2t_{1/2}) is: t1/2=ln⁡2k=0.693kt_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}

Notice that t1/2t_{1/2} is completely independent of the initial concentration of reactant [A]o[\text{A}]_o. Hence, a graph of t1/2t_{1/2} against [A]o[\text{A}]_o for a first-order reaction should be a horizontal straight line parallel to the x-axis.

The graph shown in Statement II depicts t1/2∝[A]ot_{1/2} \propto [\text{A}]_o (a straight line passing through the origin), which represents a zero-order reaction (t1/2=[A]o2kt_{1/2} = \frac{[\text{A}]_o}{2k}), not a first-order reaction.

Therefore, Statement II is false.

Conclusion:

  • Statement I is true.
  • Statement II is false.

Thus, the correct option is C.

Activation Energy Temperature Coefficient and Half Life Graph | Chemistry PYQ Solution - JEE Challenger