MCQ
A reaction rate constant is given by $k = 1.2 \times {10^{14}}{e^{ - (25000/RT)}}\,{\sec ^{ - 1}}$ It means
  • A
    $\log \,k$ versus $\log \,T$ will give a straight line with slope as $-25000$
  • B
    $\log \,k$ versus $T$ will give a straight line with slope as $-25000$
  • C
    $\log \,k$ versus $\log \,1/T$ will give a straight line with slope as $-25000$
  • $\log \,k$ versus $1/T$ will give a straight line

Answer

Correct option: D.
$\log \,k$ versus $1/T$ will give a straight line
d
(d)According to the Arrihenius equation a straight line is to be `obtained by plotting the logarithm of the rate constant of a chemical reaction ($log \,K$) against $1/T.$

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