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Q.

Variation of $\log_{10}$ K with $\frac{1}{T}$ is shown by the following graph in which straight line is at $45^{0},$ hence $∆ H^{o}$ is

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a

+4.606 cal

b

$- 4.606 cal$

c

2 cal

d

-2cal

answer is B.

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Detailed Solution

Slope of plot = $\frac{- ∆ H^{0}}{2.303 R} = 1$

$\Rightarrow ∆ H^{0} = - 2.303 \times 2 = - 4.606 cal$

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