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Q.

Graph between log k and 1/T (k = rate constant in $\sec^{- 1}$ and T is the temperature in k) is straight line. If OX = 5 and slope of line $= - \frac{1}{2.303} t h e n E_{a} i s$

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a

$2.303 \times 2 c a l$

b

$\frac{2.303}{2}$

c

$\frac{2}{2.303}$ cal

d

2 cal

answer is C.

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Detailed Solution

Slope $= \frac{- E_{a}}{2.303 R} = - \frac{1}{2.303} ∴ E a = R$

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Graph between log k and 1/T (k = rate constant in sec−1 and T is the temperature in k) is straight line. If OX = 5 and slope of line =−12.303 then Ea is