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Q.

Figure shows a graph in $\log_{10} K vs \frac{1}{T}$ where K is rate constant and T is temperature. The straight line BC has slope, $| \tan θ | = \frac{1}{2 .303}$ and an intercept of 5 on y-axis. Thus $E_{A}$ , the energy of activation is:

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a

$2 / 2.303 cal$

b

$\frac{2.303}{2}$ cal

c

$2.303 \times 2 cal$

d

$2 cal$

answer is C.

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Detailed Solution

$\log_{10} K vs . \frac{1}{T} is linear;$

$slope= \frac{- E_{A}}{2 .303 R} {intercept = log}_{10} A$

Since, $| \tan θ | = \frac{1}{2 .303}$

$∴ \frac{E_{A}}{2 .303 R} = \frac{1}{2 .303} \Rightarrow E_{a} = R = 2 cal$

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Figure shows a graph in log⁡10 K vs 1T where K is rate constant and T is temperature. The straight line BC has slope, |tan⁡θ|=12.303 and an intercept of 5 on y-axis. Thus EA, the energy of activation is: