Q54 / 180 · JIPMER 2017 · Chemistry

In the following graph. Figure The slope of line $A B$ gives the information of the

  • A
    value of $\frac{E_a}{2.303}$
  • B
    value of $\frac{2.303}{E_a}$
  • Correct answer
    C
    value of $-\frac{E_a}{2.303 R}$
  • D
    value of $-\frac{E_a}{2.303 R T}$

Solution

$\because \operatorname{In} K=\ln A-\frac{E_a}{R T}$ $\operatorname{In} K=\operatorname{In} A-\frac{E_a}{R} \cdot \frac{1}{T}$ Thus, $\log _{10} K=\log A-\frac{E_a}{2.303 R T}$ $\log K=\log A-\frac{E_a}{2.303} \cdot \frac{1}{T}$

Hence, the correct answer is option: 'C'.

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