Question

When a graph between log K and 1/T is drawn a straight line is obtained. The point at which line cuts y -axis and x -axis respectively correspond to the temp :

• A. 0, Ea / 2.303 R log A
• B. $\infty$, Ea / (R ln A)
• C. 0, log A
• D. None of these.

Answer 1

Answer: (B)

$\infty$, Ea / (R ln A)

Answer 2

$\log K = \left\lbrack \frac{- E_{a}}{2.303R} \right\rbrack \times \frac{1}{T} + \log AatYaxies\left\lbrack \frac{1}{T} \right\rbrack = 0T = \infty$

$at \times axis \log{}k = 0 \frac{- E_{a}}{2.303R} \times \frac{1}{T} = - \log A \frac{E_{a}}{RInA} = T$