CBSE Class 12 · 2001 · Delhi · Chemical KineticsShort Answer

The rate constant k of a reaction varies with temperature according to the equation

lo g k = constant - \frac{E _{a}}{2.303 R} \cdot \frac{1}{T}

, where E

_{a}

is the activation energy. When a graph is plotted for log k vs 1/T, a straight line with a slope of −6670 K is obtained. Calculate the energy of activation for this reaction. State the units. (R = 8.314 J K

^{- 1}

mol

^{- 1}

)

Answer: E_a ≈ 127.7 kJ mol⁻¹

Solution

Slope of the log k vs 1/T line:

slope = - \frac{E _{a}}{2.303 R} = - 6670 K

E_{a} = 6670 \times 2.303 \times 8.314

E_{a} = 1.277 \times 1 0^{5} J mo l^{- 1} = 127.7 kJ mo l^{- 1}

CBSE Class 12 Board · 2001 · Delhi · Subjective

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